Help with the proof of the formula for the derivative of the implicit function of two variables. My question is at the end. It's about the proof of the following theorem in my calculus textbook.
Theorem:
If an equation $F(x,y,z)=0$ determines an implicit differentiable function $f$ of two variables x and y such that $z = f(x,y)$ for every (x,y) in the domain of $f$, then $$\frac{\partial z}{\partial x} = -\frac{F_x(x,y,z)}{F_z(x,y,z)}, \frac{\partial z}{\partial y} = -\frac{F_y(x,y,z)}{F_z(x,y,z)}$$
Proof:
The statement $F(x,y,z)=0$ determines a function $f$ such that $z=f(x,y)$ means that $F(x,y,f(x,y))$ for every $(x,y)$ in the domain of $f$. Consider the composite function $F$ of $x$ and $y$ defined as follows:
$$ w = F(u,v,z), \text{where} u=x, v=y, z=f(x,y)$$
$$\frac{\partial w}{\partial x} = \frac{\partial w}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial w}{\partial v}\frac{\partial v}{\partial x} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial x}
$$
Since $w = F(x,y,f(x,y))=0$ for every x and every y, it follows that $\frac{\partial w}{\partial x} = 0$. Moreover, since $\frac{\partial u}{\partial x} = 1$ and $\frac{\partial v}{\partial x} = 0$, our chain rule formula for $\frac{\partial w}{\partial x}$ may be written as
$$ 0 = \frac{\partial w}{\partial x}(1) + \frac{\partial w}{\partial y}(0) + \frac{\partial w}{\partial z}\frac{\partial z}{\partial x}$$
My question is how is it possible in the last step that we retain $\frac{\partial w}{\partial x}$ on the RHS of the equation when we've stated earlier that $\frac{\partial w}{\partial x} = 0$ on the LHS of the equation?
 A: I think the problem is just a  notational one.
Let me briefly adopt a notation I find awkward. Let $\phi_k$ denote the derivative of $\phi$ with respect to the $k$th variable. So, using the notation above, we would have $F_2(x,y,z) = F_y(x,y,z)$.
Let $\phi(x,y) = F(x,y,f(x,y))$. Then $\phi_1(x,y) = F_1(x,y,f(x,y))+F_3(x,y,f(x,y)) f_1(x,y)$, and
$\phi_2(x,y) = F_2(x,y,f(x,y))+F_3(x,y,f(x,y)) f_2(x,y)$.
In the question above, the author has written (for the first equation, for example) the equivalent of
$\phi_1(x,y) = F_1(x,y,f(x,y))+F_2(x,y,f(x,y)) 0 + F_3(x,y,f(x,y)) f_1(x,y)$,
where the zero arises because we are differentiating with respect to $x$ instead of $y$. However, it is very confusing to include a term that arises only when differentiating with respect to another independent variable (the second variable of $F$, in this case).
Also, the author glibly switches $\frac{\partial w}{\partial u}$ to $\frac{\partial w}{\partial x}$, and similarly for $v \to y$. In one case the term $\frac{\partial w}{\partial x}$ refers to the 'total' derivative, and in the other a partial derivative. This is, at best, confusing.
Since $\phi(x,y) = 0$, we have $F_1(x,y,f(x,y))+F_3(x,y,f(x,y)) f_1(x,y) = 0$ and $F_2(x,y,f(x,y))+F_3(x,y,f(x,y)) f_2(x,y) = 0$, which gives rise to the equations
$f_1(x,y) = -\frac{F_1(x,y,f(x,y))}{F_3(x,y,f(x,y))}$, and 
$f_2(x,y) = -\frac{F_2(x,y,f(x,y))}{F_3(x,y,f(x,y))}$.
Now one can revert to the more classical notation.
