To find dy/dx we must take the derivative of the given function implicitly. Notice the term $\displaystyle x\tan^{-1}(y)$ will require the use of the Product Rule, because it is a composition of two separate functions multiplied by each other. Every other term in the given function can be derived in a straight-forward manner, but this term tends to mess with many students. Remember to use the Product Rule:

Product Rule: $\displaystyle \frac{d}{dx}[f(x)g(x)]=f(x)g'(x)+g(x)f'(x)$

Now if we take the derivative of each component of the given problem statement:

$\displaystyle 3x^2\frac{dx}{dx}+[(x)(\frac{1}{1+y^2}(\frac{dy}{dx}))+(1)(\frac{dx}{dx})(\tan^{-1}(y))]=e^y\frac{dy}{dx}$

Notice that anytime we take the derivative of a term with x involved we place a "dx/dx" next to it, but this is equal to "1".

So this now becomes:
$\displaystyle 3x^2+[\frac{x}{1+y^2}\frac{dy}{dx}+\tan^{-1}(y)}]=e^y\frac{dy}{dx}$

Now if we place all the terms with a "dy/dx" onto one side and factor out we can solved for it:

$\displaystyle 3x^2+\tan^{-1}(y)=e^y\frac{dy}{dx}-\frac{x}{1+y^2}\frac{dy}{dx}$

$\displaystyle 3x^2+\tan^{-1}(y)=\frac{dy}{dx}(e^y-\frac{x}{1+y^2})$

$\displaystyle \frac{dy}{dx}=\frac{3x^2+\tan^{-1}(y)}{e^y-\frac{x}{1+y^2}}$

This is one of the answer choices.

To find dx/dy we must take the derivative of the given function implicitly. Notice the term $\displaystyle x\tan^{-1}(y)$ will require the use of the Product Rule, because it is a composition of two separate functions multiplied by each other. Every other term in the given function can be derived in a straight-forward manner, but this term tends to mess with many students. Remember to use the Product Rule:

Product Rule: $\displaystyle \frac{d}{dx}[f(x)g(x)]=f(x)g'(x)+g(x)f'(x)$

Now if we take the derivative of each component of the given problem statement:

$\displaystyle 3x^2\frac{dx}{dy}+[(x)(\frac{1}{1+y^2}(\frac{dy}{dy}))+(1)(\frac{dx}{dy})(\tan^{-1}(y))]=e^y\frac{dy}{dy}$

Notice that anytime we take the derivative of a term with y involved we place a "dy/dy" next to it, but this is equal to "1".

So this now becomes:
$\displaystyle 3x^2\frac{dx}{dy}+[\frac{x}{1+y^2}+\tan^{-1}(y)\frac{dx}{dy}]=e^y$

Now if we place all the terms with a "dx/dy" onto one side and factor out we can solved for it:

$\displaystyle 3x^2\frac{dx}{dy}+\tan^{-1}(y)\frac{dx}{dy}=e^y-\frac{x}{1+y^2}$

$\displaystyle \frac{dx}{dy}[3x^2+\tan^{-1}(y)]=e^y-\frac{x}{1+y^2}$

$\displaystyle \frac{dx}{dy}=\frac{e^y-\frac{x}{1+y^2}}{3x^2+\tan^{-1}(y)}$

This is one of the answer choices.

We must take the derivative $\displaystyle \small \left(\small \frac{\mathrm{d} y}{\mathrm{d} x}\right)$ because that will give us the slope. On the left side we'll get 

$\displaystyle \small 3y^2\frac{\mathrm{d}y }{\mathrm{d} x}-(1/y)\frac{\mathrm{d} y}{\mathrm{d} x}$, and on the right side we'll get $\displaystyle \small 2x$.

We include the $\displaystyle \small \frac{\mathrm{d} y}{\mathrm{d} x}$ on the left side because $\displaystyle \small y$ is a function of $\displaystyle \small x$, so its derivative is unknown (hence we are trying to solve for it!).

Now we can factor out a $\displaystyle \small \frac{\mathrm{d} y}{\mathrm{d} x}$ on the left side to get 

$\displaystyle \small \small \frac{\mathrm{d} y}{\mathrm{d} x}(3y^2-1/y)=2x$ and divide by $\displaystyle \small 3y^2-1/y$ in order to solve for $\displaystyle \small \frac{\mathrm{d} y}{\mathrm{d} x}$.

Doing this gives you

 $\displaystyle \small \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2x}{3y^2-1/y}$.

We want to find the slope at $\displaystyle \small (1,1)$, so we can sub in $\displaystyle \small 1$ for $\displaystyle \small x$ and $\displaystyle \small y$. 

$\displaystyle \small \small \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2(1)}{3(1)^2-1/(1)}=2/2=1$.

To find $\displaystyle g'(x)$, substitute $\displaystyle u = x ^{2} + x$ and use the chain rule:

$\displaystyle g(x) = e ^{x^{2}+x}$

$\displaystyle g'(x) = \frac{\mathrm{d} }{\mathrm{d} x} e ^{x^{2}+x}$

$\displaystyle = \frac{\mathrm{d} }{\mathrm{d} x} e ^{u}$

$\displaystyle = \frac{\mathrm{d}u }{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u}e ^{u}$

$\displaystyle = \frac{\mathrm{d}u }{\mathrm{d} x} \cdot e ^{u}$

$\displaystyle = \frac{\mathrm{d} }{\mathrm{d} x}\left ( x ^{2} + x \right ) \cdot \frac{\mathrm{d} }{\mathrm{d} u}e ^{u}$

$\displaystyle = \left ( 2x + 1 \right ) \cdot e ^{u}$

$\displaystyle = \left ( 2x + 1 \right ) \cdot e ^{ x ^{2} + x}$

$\displaystyle g'(x)= \left ( 2x + 1 \right ) e ^{ x ^{2} + x}$

Plug in 3:

$\displaystyle g'(3)= \left ( 2 \cdot 3 + 1 \right ) \cdot e ^{ 3 ^{2} + 3}$

$\displaystyle = \left ( 6 + 1 \right ) \cdot e ^{ 9+ 3} =7e ^{ 12}$

To find $\displaystyle g'(x)$, substitute $\displaystyle u =4x - 7$ and use the chain rule: 

$\displaystyle g (x) = \frac{1}{4x - 7}$

$\displaystyle g '(x) = \frac{\mathrm{d} }{\mathrm{d} x}\cdot \frac{1}{4x - 7}$

$\displaystyle = \frac{\mathrm{d} }{\mathrm{d} x}\cdot \frac{1}{u}$

$\displaystyle = \frac{\mathrm{d} u}{\mathrm{d} x}\cdot \frac{\mathrm{d} }{\mathrm{d} u} \frac{1}{u}$

$\displaystyle = \frac{\mathrm{d} }{\mathrm{d} x} (4x-7) \cdot \frac{\mathrm{d} }{\mathrm{d} u} u ^{-1}$

$\displaystyle =4 \cdot (-1 ) u ^{-1-1}$

$\displaystyle =-4 u ^{-2} = - \frac{4}{u^{2}}= - \frac{4}{(4x-7)^{2}}$

So

$\displaystyle g'(x)= - \frac{4}{(4x-7)^{2}}$

and

$\displaystyle g'(1)= - \frac{4}{(4 \cdot 1 -7)^{2}}$

$\displaystyle = - \frac{4}{(4 -7)^{2}}$

$\displaystyle = - \frac{4}{(-3)^{2}} = - \frac{4}{9}$

To find $\displaystyle g'(x)$, substitute $\displaystyle u = x +\frac{ \pi }{4}$ and use the chain rule:

$\displaystyle g (x) = \tan \left ( x +\frac{ \pi }{4}\right )$

$\displaystyle g' (x) = \frac{\mathrm{d} }{\mathrm{d} x}\tan \left ( x +\frac{ \pi }{4}\right )$

$\displaystyle g' (x) = \frac{\mathrm{d} }{\mathrm{d} x}\tan u$

$\displaystyle =\frac{\mathrm{d} u}{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \tan u$

$\displaystyle =\frac{\mathrm{d} }{\mathrm{d} x} \left ( x +\frac{ \pi }{4} \right )\cdot \frac{\mathrm{d} }{\mathrm{d} u} \tan u$

$\displaystyle =\frac{\mathrm{d} }{\mathrm{d} x} \left ( x +\frac{ \pi }{4} \right ) \cdot \sec ^{2} u$

$\displaystyle =1 \cdot \sec ^{2}\left ( x +\frac{ \pi }{4}\right ) = \sec ^{2}\left ( x +\frac{ \pi }{4}\right )$

 $\displaystyle g'(x) = \sec ^{2}\left ( x +\frac{ \pi }{4}\right )$

$\displaystyle g'(\pi ) = \sec ^{2}\left ( \pi +\frac{ \pi }{4}\right )$

$\displaystyle g'(\pi ) = \sec ^{2}\frac{ 5 \pi }{4}$

$\displaystyle = \left ( - \sqrt{2 }\right )^{2} = 2$

To find $\displaystyle g'(x)$, substitute $\displaystyle u = x ^{2}$ and use the chain rule: 

$\displaystyle g (x) = 3 ^{x^{2}}$

$\displaystyle g ' (x) = \frac{\mathrm{d} }{\mathrm{d} x} 3 ^{x^{2}}$

$\displaystyle = \frac{\mathrm{d} }{\mathrm{d} x} 3 ^{u}$

$\displaystyle = \frac{\mathrm{d}u }{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} 3 ^{u}$

$\displaystyle = \frac{\mathrm{d}}{\mathrm{d} x} x^{2} \cdot \frac{\mathrm{d} }{\mathrm{d} u} 3 ^{u}$

$\displaystyle = 2x \cdot \ln 3 \cdot 3 ^{u}$

$\displaystyle = 2x \ln 3 \cdot 3 ^{x^2}$

So $\displaystyle g'(x)= 2x \ln 3 \cdot 3 ^{x^2}$

and

$\displaystyle g'(-1)= 2(-1) \ln 3 \cdot 3 ^{(-1)^2}$

$\displaystyle = -2 \ln 3 \cdot 3 ^{1} = -2 \ln 3 \cdot 3 = -6 \ln 3$

To find $\displaystyle g'(x)$, substitute $\displaystyle u = \pi x^{2}$ and use the chain rule:

$\displaystyle g \left (x \right )= \cos \left ( \pi x^{2}\right )$

$\displaystyle g'(x)= \frac{\mathrm{d} }{\mathrm{d} x} \cos \left ( \pi x^{2}\right )$

$\displaystyle = \frac{\mathrm{d} }{\mathrm{d} x} \cos u$

$\displaystyle =\frac{\mathrm{d} u}{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \cos u$

$\displaystyle =\frac{\mathrm{d} }{\mathrm{d} x} \pi x^{2} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \cos u$

$\displaystyle = \pi \cdot 2 x\cdot \left ( - \sin u \right )$

$\displaystyle = \pi \cdot 2 x\cdot \left [- \sin \left ( \pi x^{2} ) \right ]$

$\displaystyle =-2 \pi x \sin \left ( \pi x^{2} )$

So $\displaystyle g'(x)=-2 \pi x \sin \left ( \pi x^{2} )$

and 

$\displaystyle g ' \left ( \frac{1}{2} \right )=-2 \pi\cdot \frac{1}{2} \sin \left [\pi \cdot \left ( \frac{1}{2} \right )^{2} \right ]$

$\displaystyle =- \pi \sin \left ( \frac{\pi}{4} \right )$

$\displaystyle =- \pi \cdot \frac{\sqrt{2}}{2} = - \frac{ \pi \sqrt{2}}{2}$

Consider this function a composition of two functions, f(g(x)). In this case, f(x) is ln(x) and g(x) is 3x - 7. The derivative of ln(x) is 1/x, and the derivative of 3x - 7 is 3. The derivative is then $\displaystyle \frac{1}{3x-7} \cdot 3 = \frac{3}{3x-7}$. 

Consider this function a composition of two functions, f(g(x)). In this case, $\displaystyle f(x) =2 \sin (x)$ and $\displaystyle g(x) = x^2$. According to the chain rule, $\displaystyle f(g(x)) ' = f'(g(x))\cdot g'(x)$. Here, $\displaystyle f'(g(x)) = 2 \cos (x^2 )$ and $\displaystyle g'(x) = 2x$, so the derivative is $\displaystyle 2 \cos (x^2 ) \cdot 2x = 4x \cdot \cos(x^2)$