Why can I take $y=g(x,z)$ as a constant when doing partial differentiation? I was trying to solve this question:

Given $F(x,y,z) = 0$, and $x = f(y, z)$, $y = g(x, z)$, $z = h(x, y)$, show $\frac{\partial{x}}{\partial{y}}\frac{\partial{y}}{\partial{z}}\frac{\partial{z}}{\partial{x}} = -1$

I searched the solution, and one of the steps states to take the $\frac{\partial{F}}{\partial{x}}$ while holding $y$ constant. However, isn't $y$ defined in terms of $x$ and $z$? Therefore, when I execute this step and take $\frac{\partial{F}}{\partial{x}} = F'_{x}\frac{\partial{x}}{\partial{x}} + F'_{y}\frac{\partial{y}}{\partial{x}} + F'_{z}\frac{\partial{z}}{\partial{x}}$, then since $y$ is a function of $x$ and $z$, the $\frac{\partial{y}}{\partial{x}}$ term shouldn't be $0$ right? I know I'm wrong, in that the term should be $0$, but I just don't understand why since $y$ is defined in terms of $x$ and $z$
 A: You need to specify exactly which function you are dealing with.  Different formulas can look the same.  $F(x,y) \neq F(y,z) \neq F(x,z)$.  If
$$F(x,y,z) = 0, F(f(y,z), y, z) = 0$$
then
$$\frac{\partial F}{\partial y} = \frac{\partial F}{\partial x}\frac{\partial f}{\partial y} + \frac{\partial F}{\partial y}$$
Note that the same looking term "$\frac{\partial F}{\partial y}$" is on both sides but does not mean the same thing!  It is admittedly a pretty awful notation.  On the left we have $F(y,z)$ and on the right $F(x,y,z)$ and while people try to cope by adding subscripts explaining what is held constant, it's still confusing.
One solution is to simply make sure to give every function a different name.
Let $F(x,y,z) = F(f,y,z)$ and $f(y,z) = x$ and $G(y,z) = F(f(y,z), y, z)$ so $G$ has two inputs but all the same values as $F$ with three inputs.  Then we can say $$\frac{\partial G}{\partial y} = \frac{\partial F}{\partial x}\frac{\partial f}{\partial y} + \frac{\partial F}{\partial y}$$
