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?
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.