What is $\frac{\partial z}{\partial y}$ for $z = F(u, v, y), u = f(v, y)$ and $v = g(x, y)$?

I'm reviewing the chain rule for partial derivatives, and since I've never actually learned it other than picking it up while learning other stuff in courses, I'm not 100% sure I'm doing it correctly when it comes to more complicated interdependencies.

So my question is what $\frac{\partial z}{\partial y}$, where $z = F(u, v, y), u = f(v, y)$ and $v = g(x, y)$ would be? My understanding is that it's:

$\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial v}\frac{\partial v}{\partial y} + \frac{\partial z}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial y} + \frac{\partial z}{\partial y}$

Is this correct? I'm unsure about the first two terms and whether the third one would somehow be subsumed in the first one.

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You might want to be more verbose with your notation. Partial derivatives act on functions. So instead of writing $(\partial/\partial y) z$ on both sides of the equation, try writing $(\partial/\partial y) F(u, v, y)$ versus $(\partial/\partial y) F(x, y)$. That should clear things up. –  Chris Culter Jul 22 '13 at 18:20
@ChristopherCulter, you're right, of course, but I did this quickly and figured it was clear enough what I was after, so I didn't want to spend too much time on writing it up nicely in TeX. –  Ryker Jul 22 '13 at 19:06

It is always been helpful for me, to think of this as a tree. Then you walk to the leaves that contain to variable of your interest.

Derivative with respect to $x$ would be: $\frac{\partial F}{\partial x} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial v} \frac{\partial v}{\partial x} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial x}$

Derivative with respect to $y$ would be: $\frac{\partial F}{\partial y} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial v} \frac{\partial v}{\partial y} + \frac{\partial z}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial y} + \frac{\partial z}{\partial y}$

Damn, I noticed I put the right thing in the thread title and my solution, but not the question itself (I edited it now). Basically, I was asking for $\frac{\partial z}{\partial y}$, not $\frac{\partial z}{\partial z}$. Could you perhaps verify if my solution is correct? Sorry about the confusion. –  Ryker Jul 22 '13 at 18:39
For now I've only up-voted it, because it is helpful, but it doesn't actually answer the question. I feel that's how it's done on this website, so if you change your answer to the solution for $\frac{\partial z}{\partial y}$, I'll be glad to accept it! –  Ryker Jul 22 '13 at 21:37