To find $\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x}$ we must use implicit differentiation, which is an application of the chain rule.

Taking $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}$ of both sides of the equation, we get

$\displaystyle 2x+x^2\frac{\mathrm{d} y}{\mathrm{d} x}+2y\frac{\mathrm{d} y}{\mathrm{d} x}=1$

using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}ax=a$

Note that for every derivative of a function with y, the additional term $\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}$ appears; this is because of the chain rule, where $\displaystyle y = g(x)$, so to speak, for the function it appears in. 

Using algebra to rearrange, we get

$\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1-2x}{2y+x^2}$

The derivative of the function is equal to

$\displaystyle f'(x)=x\sec^2(x^2)\sqrt{e^\tan(x^2)}$

and was found using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Before simplification, the derivative we get is

$\displaystyle f'(x)=\frac{e^{\tan(x^2)^{-\frac{1}{2}}}}{2}(2x\sec^2(x^2)e^{\tan(x^2)})$

Note that the square root, the exponential, and the tangent function all utilize chain rule when taking their derivatives. 

The derivative of the function is equal to

$\displaystyle f'=-2e^{2x}\sin(e^{2x})$

and was found using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Note that the first rule - the chain rule - was used three times for the function: the cosine, contained the exponential which itself was raised to a function. (Note that sometimes the exponential rule is written as $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{d} u}{\mathrm{d} x}$, which itself is the chain rule.)

The derivative of the function is equal to

$\displaystyle \kappa'(T)=\frac{AR}{T^2}e^{-\frac{R}{T}}$

and was found using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

The chain rule was used for the function contained in the exponential function (or, as written as a rule, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{d} u}{\mathrm{d} x}$).

The equation given is not written and $\displaystyle y = f(x)$. Instead, it is written with $\displaystyle y$'s and $\displaystyle x$'s on the same side of the equation. This suggests we should try implicit differentiation, which means find the derivative of both sides with respect to $\displaystyle x$.

"With respect to $\displaystyle x$" means that we treat every other variable as a function of $\displaystyle x$. So $\displaystyle y = f(x)$, and the derivative of $\displaystyle y$ is a chain rule. This will be emphasized later in the explanation.

First we must differentiate both sides with respect to $\displaystyle x$.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}[y^2 + 2x^2y - \frac{5x}{y}] = \frac{\mathrm{d} }{\mathrm{d} x}[2]$

We have multiple terms on the left hand side, so we will differentiate each term individually.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}[y^2] + \frac{\mathrm{d} }{\mathrm{d} x}[2x^2y] + \frac{\mathrm{d} }{\mathrm{d} x}[\frac{-5x}{y}] = \frac{\mathrm{d} }{\mathrm{d} x}[2]$

For the first term, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}[y^2]$, we will use the power rule and also use the chain rule, since we must assume that $\displaystyle y = f(x)$.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}[y^2]$

$\displaystyle {\color{Blue} 2[y]^{2-1}}{\color{DarkRed} (\frac{\mathrm{d} y}{\mathrm{d} x})}$

$\displaystyle {\color{Blue} 2y}{\color{DarkRed} \frac{\mathrm{d} y}{\mathrm{d} x}}$

The blue part is the power rule. The red is from the chain rule. The red part is the derivative of y with respect to x, which is currently unknown. Remember that y is some unknown function of x, whose derivative is also unknown. We can only write $\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}$ for the derivative of y.

Now we find the second term's derivative,

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}[{\color{Blue} 2x^2}{\color{DarkRed} y}]$

This is a product rule. To help with the product rule, the two pieces are color coded. Remember that the power rule is $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}[{\color{Blue} f(x)}\cdot{\color{DarkRed} g(x)}] ={\color{Blue} f(x)}\cdot {\color{DarkRed} g'(x)} + {\color{Blue} f'(x)} \cdot {\color{DarkRed} g(x)}$

Applying this, we get

$\displaystyle {\color{Blue} (2x^2)}{\color{DarkRed} (\frac{\mathrm{d} y}{\mathrm{d} x})} + {\color{Blue} (4x)}{\color{DarkRed} (y)}$

Simplifying gives us

$\displaystyle 2x^2 \cdot \frac{\mathrm{d} y}{\mathrm{d} x} + 4xy$

For the third term, $\displaystyle \frac{-5x}{y}$, we will algebraically rewrite it as $\displaystyle -5xy^{-1}$, so we can apply the product rule instead of the quotient rule. This is just a personal preference, The quotient rule would work as well.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}[{\color{Blue} -5x}{\color{DarkRed} y^{-1}}]$

$\displaystyle {\color{Blue} (-5x)}{\color{DarkRed} (-1y^{-1-1})(\frac{\mathrm{d} y}{\mathrm{d} x})} + {\color{Blue} (-5)}{\color{DarkRed} (y^{-1})}$

remember that the derivative of $\displaystyle {\color{DarkRed} y^{-1}}$ with respect to $\displaystyle x$ requires the chain rule, resulting in the $\displaystyle {\color{DarkRed} (\frac{\mathrm{d} y}{\mathrm{d} x})}$.Simplifying gives us

$\displaystyle 5xy^{-2}\frac{\mathrm{d} y}{\mathrm{d} x} - 5y^{-1}$

The right hand side of the equation is a constant, so its derivative is zero.

Assembling all the parts back together, we have

$\displaystyle 2y\frac{\mathrm{d} y}{\mathrm{d} x} + 2x^2\frac{\mathrm{d} y}{\mathrm{d} x} + 4xy + 5xy^{-2}\frac{\mathrm{d} y}{\mathrm{d} x} - 5y^{-1 }=0$

Now that we have differentiated the equation, we need to algebraically solve for $\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}$.

First, we should move all terms with a $\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}$ to one side, which in this case is already done. Then we should move all terms without a $\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}$ to the opposite side of the equation. Doing so, we get

$\displaystyle 2y\frac{\mathrm{d} y}{\mathrm{d} x} + 2x^2\frac{\mathrm{d} y}{\mathrm{d} x} + 5xy^{-2}\frac{\mathrm{d} y}{\mathrm{d} x} = -4xy+ 5y^{-1}$

Then we will will factor out the common factor of $\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}$. This results in

$\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}[2y + 2x^2 + 5xy^{-2}] = -4xy + 5y^{-1}$

Then, to isolate $\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}$, we divide both sides by $\displaystyle [2y + 2x^2 + 5xy^{-2}]$.

$\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-4xy + 5y^{-1}}{2y + 2x^2 + 5xy^{-2}}$

Now we need to simplify and get rid of the negative exponents. To do this, we can simply multiply the numerator and denominator by $\displaystyle y^2$. This will result in

$\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-4xy + 5y^{-1}}{2y + 2x^2 + 5xy^{-2}} \cdot \frac{y^2}{y^2}$

$\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-4xy^3 + 5y}{2y^3 + 2x^2y^2 + 5x}$

Which is the correct answer.

Given the function $\displaystyle \small \small \small z(x)=\cos x^{12}$, we can find its derivative using the chain rule, which states that 

$\displaystyle \small [f(g(x))]'=f'(g(x))g'(x)$

where $\displaystyle \small \small f(x)=\cos x$ and $\displaystyle \small \small g(x)=x^{12}$  for $\displaystyle \small z(x)$. We have $\displaystyle \small f'(x)=-\sin x$ and $\displaystyle \small g'(x)=12x^{11}$, which gives us 

$\displaystyle \small \small z'(x)=-\sin x^{12}\cdot 12x^{11}=-12x^{11}\sin x^{12}$

Given the function $\displaystyle \small \small \small \small z(x)=e^{x^{-1}}$, we can find its derivative using the chain rule, which states that 

$\displaystyle \small [f(g(x))]'=f'(g(x))g'(x)$

where $\displaystyle \small \small \small f(x)=e^{x}$ and $\displaystyle \small \small \small g(x)=x^{-1}$  for $\displaystyle \small z(x)$. We have $\displaystyle \small \small f'(x)=e^x$ and $\displaystyle \small \small g'(x)=-x^{-2}$, which gives us 

$\displaystyle \small \small \small z'(x)=e^{x^{-1}}\cdot (-x^{-2})=-\frac{e^{x^{-1}}}{x^2}$

Given the function $\displaystyle \small \small \small z(x)=e^{-x+x^3}$, we can find its derivative using the chain rule, which states that 

$\displaystyle \small [f(g(x))]'=f'(g(x))g'(x)$

where $\displaystyle \small \small \small f(x)=e^{x}$ and $\displaystyle \small \small \small \small g(x)=-x+x^3$  for $\displaystyle \small z(x)$. We have $\displaystyle \small \small f'(x)=e^x$ and $\displaystyle \small \small \small g'(x)=-1+3x^2$, which gives us 

$\displaystyle \small \small \small \small \small z'(x)=e^{-x+x^3}\cdot (-1+3x^{2})$

Given the function $\displaystyle \small \small \small \small \small z(x)=e^{\cos x+\sin x}$, we can find its derivative using the chain rule, which states that 

$\displaystyle \small [f(g(x))]'=f'(g(x))g'(x)$

where $\displaystyle \small \small \small \small f(x)=e^{x}$ and $\displaystyle \small \small \small \small \small \small g(x)=\cos x+\sin x$  for $\displaystyle \small z(x)$. We have $\displaystyle \small \small f'(x)=e^x$ and $\displaystyle \small \small \small \small \small \small g'(x)=-\sin x+\cos x$, which gives us 

$\displaystyle \small \small z'(x)=e^{\cos x+\sin x}\cdot (-\sin x+\cos x)$

Given the function $\displaystyle \small \small \small \small z(x)=e^{e^x-x^5}$, we can find its derivative using the chain rule, which states that 

$\displaystyle \small [f(g(x))]'=f'(g(x))g'(x)$

where $\displaystyle \small \small \small \small f(x)=e^{x}$ and $\displaystyle \small \small \small \small \small g(x)=e^x-x^5$  for $\displaystyle \small z(x)$. We have $\displaystyle \small \small f'(x)=e^x$ and $\displaystyle \small \small \small \small g'(x)=e^x-5x^4$, which gives us 

$\displaystyle \small \small \small \small \small \small \small z'(x)=e^{e^x-x^5}\cdot (e^x-5x^4)$