Not following the derivation of $\frac{dy}{dx}=-\frac{F_x}{F_y}$ I've seen similar questions to mine asked on the forum, but I haven't seen answers that address the part I'm confused about.
My calculus textbook (Thomas from Pearson) derives the following formula to "take some of the algebra out of implicit differentiation":

Suppose the function $F(x,y)$ is differentiable and the equation $F(x,y)=0$ defines $y$ implicitly as a differentiable function of $x$. Then at any point where $F_y\neq 0$, we have  $$\frac{dy}{dx}=-\frac{F_x}{F_y}$$.

(The formula itself is pretty intuitive to me, except for the negative sign.) I feel like I am misinterpreting the derivation given, as it seems to be using $F(x,y)$ to denote two different functions and treating them as if they are the same. The derivation goes like this:

Suppose that (1) the function $F(x,y)$ is differentiable and that (2) the equation $F(x,y)=0$ defines $y$ implicitly as a differentiable function of $x$. Since $w=F(x,y)=0$, the derivative $\frac{dw}{dx}$ must be zero.

As I understand this, they are defining a new function $w:\{(x,y):F(x,y)=0\}\rightarrow\{0\}$, a level curve of the original $F(x,y)$, which is zero everywhere on its domain, and we're to suppose that its domain defines $y$ implicitly in terms of $x$. But then they continue:

... Computing the derivative [of the equation $w=F(x,y)=0$] from the chain rule, we find $$0=\frac{dw}{dx}=F_x\frac{dx}{dx}+F_y\frac{dy}{dx}=F_x+F_y\frac{dy}{dx}.$$ Therefore, we have $$\frac{dy}{dx}=-\frac{F_x}{F_y}.$$

This is where I get confused. In the example questions, it is clear that $F_x$ and $F_y$ denote the partial derivatives of the original function $F(x,y)$ of which $w$ is a level curve. But this use of the chain rule seems to assume that those are also the partials of w (which is a constant function, and should have zero derivatives, no?). I'm interpreting this as a special case of
$$\frac{dw}{dt}=\frac{\partial w}{\partial x}\frac{dx}{dt}+\frac{\partial w}{\partial y}\frac{dy}{dt}$$ where $t=x$, and where $\frac{\partial w}{\partial x}$ and $\frac{\partial w}{\partial y}$ are written as $F_x$ and $F_y$. But I'm not seeing how the former and the latter partials are equivalent. Why can we assume both that $\frac{dw}{dx}=0$ and that $F_x=\frac{\partial w}{\partial x}$, when $F_x$ is not zero in general? Or is that assumption not actually being made by using the chain rule this way? What am I missing or getting wrong here? I'd really appreciate if someone would set me on the right track so that I can get some intuition for why this theorem works.
Thanks!
 A: Yes, there are several abuses of notation here. What is happening is you're first given a smooth function $F:\Bbb{R}^2\to\Bbb{R}$; for simplicity assume that at every point $p\in\Bbb{R}^2$, we have $\frac{\partial F}{\partial y}(p)\neq 0$. The implicit function theorem tells us that if you fix such a point $p=(a,b)$, then you can find some smooth function $\eta:I\subset\Bbb{R}\to\Bbb{R}$ such that $\eta(a)=b$ and for all $t\in I$, we have $F(t,\eta(t))=0$. So, the function $w:I\to\Bbb{R}$ defined as $w(t)=F(t,\eta(t))$ is smooth and is zero at every point; i.e is the constant zero function. So, we obviously have that $w'=0$. But now what does the chain rule tell us (note that $w$ is the composition of $F$ with the function $t\mapsto (t,\eta(t))$, so chain rule is indeed the way to go)? It tells us for each $t\in I$,
\begin{align}
0&=w'(t)=\frac{\partial F}{\partial x}\bigg|_{(t,\eta(t))} \cdot 1+\frac{\partial F}{\partial y}\bigg|_{(t,\eta(t))}\cdot \eta'(t)
\end{align}
Rearranging this equation, we get
\begin{align}
\eta'(t)&=-\frac{\frac{\partial F}{\partial x}\bigg|_{(t,\eta(t))}}{\frac{\partial F}{\partial y}\bigg|_{(t,\eta(t))}}.
\end{align}
Hopefully with the different notation, it's clear what the different functions are, and how the chain rule is being applied, and where everything is evaluated.

If the $x,y$ are confusing (and I believe they are), you can write the chain rule computation as follows: for each $t\in I$,
\begin{align}
0&=w'(t)=(\partial_1F)_{(t,\eta(t))}\cdot 1+(\partial_2F)_{(t,\eta(t))}\cdot \eta'(t),
\end{align}
and hence
\begin{align}
\eta'(t)&=-\frac{(\partial_1F)_{(t,\eta(t))}}{(\partial_2F)_{(t,\eta(t))}}
\end{align}
It is an abuse of notation to use $y$ to refer to both the coordinate, and also the name of the implicitly defined function, and to use $F$ as both the original function, and the new composed function $w$, but unfortunately, it is standard practice.
A: Let's give the new function a name: $Y$.  Thus we are assuming there is a point $(x_0, y_0)$ where $F(x_0, y_0) = 0$ and $F_y(x_0, y_0) \ne 0$, and a differentiable  function $Y$ defined in a neighbourhood of $x_0$ such that $F(x, Y(x)) = 0$ in this neighbourhood, with $Y(x_0) = y_0$.  Then we can differentiate both sides of the equation $F(x, Y(x)) = 0$ using the chain rule, obtaining
$$ F_x(x, Y(x)) + F_y(x, Y(x)) Y'(x) = 0 $$
so that
$$ Y'(x) = - \frac{F_x(x, Y(x))}{F_y(x, Y(x))} $$
A: 
$$\frac{dw}{dt}=\frac{\partial w}{\partial x}\frac{dx}{dt}+\frac{\partial w}{\partial y}\frac{dy}{dt}$$ where $t=x$, and where $\frac{\partial w}{\partial x}$ and $\frac{\partial w}{\partial y}$ are written as $F_x$ and $F_y$.

Then $\dfrac{\mathrm d w}{\mathrm d t}=0$ , $\dfrac{\mathrm d x}{\mathrm d t}=1$ , $\dfrac{\mathrm d y}{\mathrm d t}=\dfrac{\mathrm dy}{\mathrm d x}$ , and so so:$$0=F_x\cdot 1+F_y\cdot\dfrac{\mathrm d y}{\mathrm d x}$$

Why can we assume both that $\mathrm dw/\mathrm dx=0$ and that $F_x=∂w/∂x$, when $F_x$ is not zero in general?

We do not assume $\mathrm dw/\mathrm dx=0$, rather it is true because $w$ is established as a constant function ($w=0$).
However, $w$ is also a function in two variables, one which is $x$ and the other $y$.   $w=F(x,y)$.   Therefore, when the partial derivative with respect to the first argument is not zero, then $y$ must be related to $x$ in some manner to make make $w$ a constant function, and that relation makes $$\dfrac{\mathrm d y}{\mathrm d x}=-~\dfrac{F_x(x,y)}{F_y(x,y)}$$

Alternatively:  Let $\partial_n F(\cdots)$ be the partial derivative with respect to the $n^{\rm th}$ argument of the function $F$.
Let $F(x,h(x))=0$ , so by the chain rule:
$$\begin{align}\dfrac{\mathrm d ~~}{\mathrm d x}F(x,h(x))&= \dfrac{\mathrm d x}{\mathrm dx}~\partial_1F(x,h(x))+\dfrac{\mathrm d h(x)}{\mathrm d x}~\partial_2F(x,h(x))\\[2ex]\therefore\qquad\dfrac{\mathrm d h(x)}{\mathrm d x}&=-\,\dfrac{\partial_1 F(x,h(x))}{\partial_2 F(x,h(x))}\end{align}$$
So let $y=h(x)$ ...

