My question that in multivariable calculus the implicit function theorem states that:
if $F(x,y)$ and $y=f(x)$, $$\frac{dy}{dx}=-\frac{\frac{\partial }{\partial x}\left(F\right)}{\frac{\partial }{\partial y}\left(F\right)}$$
I have a $2$ problems with this
$1-$ $F(x,y)$ $and$ $y=f(x)$ so his means that the function $F$ is a function of one variable which is $x$
$2-$ while we were computing $partial$ $derivatives$ we treated $y$ and $x$ as two independent variables although that $y$ changes as $x$ changes but while doing the $partial$ $derivatives$ w.r.t $x$ we treated $y$ and $x$ as two independent varaibles and considered $y$ as a constant
The function $F$ is by definition a function of two variables $x$ and $y$. It is trivial that we can take two partial derivatives of it.
This $F$ implicitly defines a function $f(x)$ of one variable by the constraint $F(x,y)=c$ where $c$ is a constant.
The correct equation for $f'(x)$ you obtain from the chain rule: $$ \frac{d}{dx}F(x,f(x))=\frac{\partial}{\partial x}F(x,f(x))+\frac{\partial}{\partial y}F(x,f(x))f'(x). $$ Now observe that the LHS of this is zero because we said $F(x,f(x))$ is constant. Can you solve this equation now for $$ f'(x)=\frac{dy}{dx}\quad ? $$