\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(2(0)y^{3} - 5(0)^{4} + 3y^{4})}{((0)^{4} + 2y^{4})}=\frac{-3y^{4}}{2y^{4}}\\&\lim_{(0,y)\rightarrow(0+,0+)}\frac{-3y^{4}}{2y^{4}}=-\frac{3}{2}\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}-\frac{(2x(x)^{3} - 5x^{4} + 3(x)^{4})}{(x^{4} + 2(x)^{4})}=0\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(26(0)^{3}y^{2} - 14(0)^{5})}{(18(0)y^{4} - 5(0)^{4}y + 3(0)^{5} + 3y^{5})}=\frac{0}{3y^{5}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{3y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}\frac{(26x^{3}(x)^{2} - 14x^{5})}{(18x(x)^{4} - 5x^{4}(x) + 3x^{5} + 3(x)^{5})}=\frac{12}{19}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

Since we won't have any divided by zero problems, we can just evaluate at the point.

\(\displaystyle \lim_{(x,y)\rightarrow (0,1)} \frac{x^2+xy}{y}=\frac{0^2+0\cdot1}{1}=0\)

Since we won't have any divided by zero problems, we can just evaluate at the point.

\(\displaystyle \lim_{(x,y)\rightarrow (0,1)} \frac{x^2+xy}{y}=\frac{0^2+0\cdot1}{1}=0\)

All we need to do is use the formula for multivariable chain rule.

\(\displaystyle \frac{dz}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}\)

\(\displaystyle \frac{\partial f}{\partial x}=2e^{xy}+2xye^{xy}\)

\(\displaystyle \frac{dx}{dt}=3t^2\)

\(\displaystyle \frac{\partial f}{\partial y}=2x^2e^{xy}\)

\(\displaystyle \frac{dy}{dt}=-2t^{-3}\)

When we put this all together, we get.

\(\displaystyle \frac{dz}{dt}=(6t^2e^{xy}+2xye^{xy})-(4t^{-3}x^2e^{xy})\)

Expand the equation for chain rule.

\(\displaystyle \frac{\partial a}{\partial x} = \frac{\partial a}{\partial x}\frac{\partial x}{\partial r}+ \frac{\partial a}{\partial y}\frac{\partial y}{\partial r} + \frac{\partial a}{\partial z}\frac{\partial z}{\partial r}\)

Evaluate each partial derivative.

\(\displaystyle \frac{\partial a}{\partial x} = 1\),  \(\displaystyle \frac{\partial x}{\partial r} =3r^2\)

\(\displaystyle \frac{\partial a}{\partial y}= -3\), \(\displaystyle \frac{\partial y}{\partial r} = 2\)

\(\displaystyle \frac{\partial a}{\partial z} =2z\), \(\displaystyle \frac{\partial z}{\partial r} = 6\)

Substitute the terms in the equation.

\(\displaystyle \frac{\partial a}{\partial x} =(1)(3r^2)+(-3)(2)+(2z)(6) = 3r^2-6+12z\)

Substitute \(\displaystyle z=6r\).

\(\displaystyle 3r^2-6+12(6r) = 3r^2+72r-6\)

The answer is:  \(\displaystyle 3r^2+72r-6\)

The chain rule states \(\displaystyle \frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}\).

Since \(\displaystyle z\) and \(\displaystyle y\) are both functions of \(\displaystyle t\), \(\displaystyle \frac{dr}{dt}\) must be found using the chain rule.

In this problem

\(\displaystyle \frac{dr}{dt}=\frac{\partial r}{\partial z}\frac{dz}{dt}+\frac{\partial r}{\partial y}\frac{dy}{dt}\)

\(\displaystyle =\left ( 2z\right )\left ( 2t\right )+\left ( -2\right )\left ( 3t^2-2\right )\)

\(\displaystyle =\left ( 2t^2\right )\left ( 2t\right )+\left ( -2\right )\left ( 3t^2-2\right )\)

\(\displaystyle =4t^3-6t^2+4\)

The chain rule states \(\displaystyle \frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}\).

Since \(\displaystyle a\) and \(\displaystyle b\) are both functions of \(\displaystyle t\), \(\displaystyle \frac{dr}{dt}\) must be found using the chain rule.

In this problem

\(\displaystyle \frac{dr}{dt}=\frac{\partial r}{\partial a}\frac{da}{dt}+\frac{\partial r}{\partial b}\frac{db}{dt}\)

\(\displaystyle =\left ( 12a^3\right )\left ( 2t-6\right )+\left ( 10b\right )\left ( 1\right )\)

\(\displaystyle =\left ( 12\left ( t^2-6t+12\right )^3\right )\left ( 2t-6\right )+\left ( 10\left ( t-5\right )\right )\left ( 1\right )\)

\(\displaystyle =\left12 ( t^6-18t^5+144t^4-648t^3+1728t^2-2592t+1728\right )\left ( 2t-6\right )+10t-50\)

\(\displaystyle =24t^7-504t^6+4752t^5-25920t^4+88128t^3-186624t^2+228096t-124416 +10t-50\)

\(\displaystyle =24t^7-504t^6+4752t^5-25920t^4+88128t^3-186624t^2+228106t-124466\)

The chain rule states \(\displaystyle \frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}\).

Since \(\displaystyle a\) and \(\displaystyle b\) are both functions of \(\displaystyle t\), \(\displaystyle \frac{dr}{dt}\) must be found using the chain rule.

In this problem

\(\displaystyle \frac{dr}{dt}=\frac{\partial r}{\partial a}\frac{da}{dt}+\frac{\partial r}{\partial b}\frac{db}{dt}\)

\(\displaystyle =b*cos(ab)(3t^2+7)+a*cos(ab)(10t-12)\)

\(\displaystyle =-(5t^2-12t+1)sin(5t^5-12t^4+36t^3-84t^2+7t)\left ( 3t^2+7\right )-(t^3+7t)sin(5t^5-12t^4+36t^3-84t^2+7t)\left ( 10t-12\right )\)

\(\displaystyle =-(15^4-36t^3+38t^2-84t+7)sin(5t^5-12t^4+36t^3-84t^2+7t)-(10t^4-12t^3+70t^2-84t)sin(5t^5-12t^4+36t^3-84t^2+7t)\)

\(\displaystyle =(-25t^4+48t^3-108t^2+168t-7)sin(5t^5-12t^4+36t^3-84t^2+7t)\)

 The chain rule states \(\displaystyle \frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}\).

Since \(\displaystyle a\) and \(\displaystyle b\) are both functions of \(\displaystyle t\), \(\displaystyle \frac{dr}{dt}\) must be found using the chain rule.

In this problem

\(\displaystyle \frac{dr}{dt}=\frac{\partial r}{\partial a}\frac{da}{dt}+\frac{\partial r}{\partial b}\frac{db}{dt}\)

\(\displaystyle \frac{\partial r}{\partial a}=cos(ab^2)(b^2)\)

\(\displaystyle =cos(405t^7-81t^6)(81t^6)\)

\(\displaystyle \frac{da}{dt}=5\)

\(\displaystyle \frac{\partial r}{\partial b}=cos(ab^2)(2ab)\)

\(\displaystyle =cos(405t^7-81t^6)(-90t^4+18t^3)\)

\(\displaystyle \frac{db}{dt}=-27t^2\)

\(\displaystyle \frac{dr}{dt}=\left ( cos(405t^7-81t^6)(81t^6) \right )(5)+(cos(405t^7-81t^6)(-90t^4+18t^3))(-27t^2)\)

\(\displaystyle =405t^6*cos(405t^7-81t^6)+(2430t^6-486t^5)cos(405t^7-81t^6)\)

\(\displaystyle =(2745t^6-486t^5)cos(405t^7-81t^6)\)