Deriving $\frac{\partial{v}}{\partial{z}}$ using chain rule A problem from one of past exams at my uni reads: 


  Suppose $xy^2 + yz^2 = 2$ and $v = xy^3$ where $x, y, z$ and $v$ are all variables. Using a chain rule, express $\partial{v}/\partial{z}$ as a function of $x, y, z$. 


I just don't think I really get multivariable calculus... I know (?) I can write $$\frac{\partial v}{\partial z}= \frac{\partial v}{\partial x}\frac{\partial x}{\partial z}+\frac{\partial v}{\partial y}\frac{\partial y}{\partial z}$$, and clearly $\frac{\partial v}{\partial x}=y^3$ and $\frac{\partial v}{\partial y}=3xy^2$, so that $$\frac{\partial v}{\partial z}= y^3\frac{\partial x}{\partial z}+3xy^2\frac{\partial y}{\partial z}.$$ Now, $xy^2+yz^2=2$, so $x=\frac{2}{y^2}-\frac{z^2}{y}$, and I probably can write $$\frac{\partial x}{\partial z} = -2\frac{z}{y}, $$ but how do I go about deriving $\frac{\partial y}{\partial z}$? Or does all of it make no sense and I should proceed in a completely different way?
We have that $xy^2+yz^2 = 2$. Then,

$$\dfrac{\partial}{\partial z} (xy^2+yz^2) = \dfrac{\partial}{\partial z}2 =0$$

Taking the partial derivative will yield

$$\begin{align}
\dfrac{\partial}{\partial z} (xy^2+yz^2) &amp;=y^2\dfrac{\partial x}{\partial z}+x\dfrac{\partial}{\partial z}y^2 + z^2\dfrac{\partial y}{\partial z}+y\dfrac{\partial}{\partial z}z^2
\\
&amp;=-2zy+2xy\dfrac{\partial y}{\partial z} + z^2\dfrac{\partial y}{\partial z}+2zy
\\
&amp;=0
\end{align}$$

Thus, we have that $$\dfrac{\partial y}{\partial z}=0$$
We have that $xy^2+yz^2 = 2$. Then,

$$\dfrac{\partial}{\partial z} (xy^2+yz^2) = \dfrac{\partial}{\partial z}2 =0$$

Taking the partial derivative will yield

$$\begin{align}
\dfrac{\partial}{\partial z} (xy^2+yz^2) &amp;=y^2\dfrac{\partial x}{\partial z}+x\dfrac{\partial}{\partial z}y^2 + z^2\dfrac{\partial y}{\partial z}+y\dfrac{\partial}{\partial z}z^2
\\
&amp;=-2zy+2xy\dfrac{\partial y}{\partial z} + z^2\dfrac{\partial y}{\partial z}+2zy
\\
&amp;=0
\end{align}$$

Thus, we have that $$\dfrac{\partial y}{\partial z}=0$$