I'm trying to understand the derivation of the formula for $\frac{\partial z}{\partial x}|_y$ given $f(x,y,z)=c$ for $c$ constant, and writing $z$ as a function of $x$ and $y$ and using the chain rule I can get to $$\frac{\partial f}{\partial x}|_y = \frac{\partial f}{\partial x}|_{yz} + \frac{\partial f}{\partial z}|_{xy} \frac{\partial z}{\partial x}|_y .$$
In order to get the formula I need to set the LHS of the equation to zero, but I'm unsure why this is the case.
Please can somebody explain what $\frac{\partial f}{\partial x}|_y$ represents for $f$, and why it is $0$ in this case?
Edit: Where $\frac{\partial z}{\partial x}|_y$ means the derivative with y held constant.
You are looking at the relation $f(x,y,z)=c$ as defining $z$ as a function of the independent variables $x,y$. Therefore, when you take the derivative with respect to $x$, $$ \partial f/\partial x+\partial f/\partial y\cdot\partial y/\partial x+\partial f/\partial z\cdot\partial z/\partial x=0 $$ you set $\partial y/\partial x=0$.
It is actually better to use differentials: $$ \partial f/\partial x\cdot dx+\partial f/\partial y\cdot dy+\partial f/\partial z\cdot dz=0 $$ and since you are considering the variation with respect to $x$, you set $dy=0$ and compute $\partial z/\partial x$.