Differentiating with respect to x and y I'm reading a proof and I'm struggling with basic calculus here. 
Given the equation, 
$F[x, F(y, z)] = F[F(x, y), z] $
set  $u = F(x, y)$ and $v = F(y, z)$
so you have 
$$F(x, v) = F(u, z)$$
Now differentiate with respect to $x$ and $y$ leads to the following 


*

*$F_{x}(x,v) = F_{x}(u,z)F_{x}(x,y)$

*$F_{y}(x,v)F_{x}(y,z) = F_{x}(u,z)F_{y}(x,y)$


I think, according to the chain rule, the derivative of  $F[x, F(y, z)]$ or $F[x, u]$ with respect to x should equal $F_{x}(u,z)F_{x}(x,y)$. So number 1 makes sense. But when you differentiate that with respect to $y$, I don't understand how you get 2. 
 A: You seem to have a typo; I presume you meant $F(x,v)$ where it says $F[x,u]$?
You don't get 2. by differenitating 1. with respect to $y$, but by differentiating the original equation, $F(x,v)=F(u,z)$, with respect to $y$.
A: You are given that $$F(x,F(y,z)) = F(F(x,y),z)$$
Calling $F(x,y) = u$ and $F(y,z) = v$, we then have that $$F(x,v) = F(u,z)$$
Differentiating $F(x,v) = F(u,z)$ with respect to $x$, gives us
$$\dfrac{\partial F(x,v)}{\partial x} + \dfrac{\partial F(x,v)}{\partial v} \underbrace{\dfrac{\partial v}{\partial x}}_{0} = \dfrac{\partial F(u,z)}{\partial u} \dfrac{\partial u}{\partial x} + \dfrac{\partial F(u,z)}{\partial z} \underbrace{\dfrac{\partial z}{\partial x}}_{0}$$
Hence, you get the first equation i.e. $$F_x(x,v) = F_u(u,z)u_x = F_u(u,z)F_x(x,y)$$
Differentiating $F(x,v) = F(u,z)$ with respect to $y$, gives us
$$\underbrace{\dfrac{\partial F(x,v)}{\partial y}}_{0} + \dfrac{\partial F(x,v)}{\partial v} \dfrac{\partial v}{\partial y} = \dfrac{\partial F(u,z)}{\partial u} \dfrac{\partial u}{\partial y} + \dfrac{\partial F(u,z)}{\partial z} \underbrace{\dfrac{\partial z}{\partial y}}_{0}$$
Hence, you get the second equation i.e. $$F_v(x,v)v_y = F_u(u,z)u_y$$
$$F_v(x,v)F_y(y,z) = F_u(u,z)F_y(x,y)$$
