Let $z = z(x,y)$ be defined implicitly by $F(x, y, z(x,y)) = 0$, where $F$ is a given function of three variables.
Prove that if $z(x,y)$ and $F$ are differentiable, then
$$\frac{dz}{dx} = - \large{\frac{\frac{dF}{dx}} {\frac {dF}{dz}}}$$
if $dF/dz \not= 0$
I am kind of confused with $z = z(x,y)$. Should I differentiate $dz/dx$ and $dz/dy$, but there's nothing inside $x$ and $y$ so what should I differentiate $dx$ and $dy$ with then? I've been told that this is something related to implicit differentiation. In some other examples, they do use differentiate $F$ i.e. $dF$ but in most examples, I only see that they are differentiating the variables inside it.
Consider the function $\varphi \, : \, x \, \longmapsto \, F(x,y,z(x,y))$. We have :
$$ \varphi(x) = 0 \tag{$\star$} $$
Differentiate (using the chain rule) this equality with respect to $x$. You will get :
$$ \varphi'(x) = \frac{\partial F}{\partial x}\big( x,y,z(x,y) \big) + \frac{\partial z}{\partial x}(x,y) \frac{\partial F}{\partial z}\big( x,y,z(x,y) \big) = 0. $$
As a consequence, if $\displaystyle \frac{\partial F}{\partial z}\big( x,y,z(x,y) \big) \neq 0$, we have :
$$ \frac{\partial z}{\partial x}(x,y) = - \frac{ \frac{\partial F}{\partial x}\big( x,y,z(x,y) \big) }{ \frac{\partial F}{\partial z}\big( x,y,z(x,y) \big) }. $$