Proof on showing if F(x,y,z)=0 then product of partial derivatives (evaluated at an assigned coordinate) is -1

The task is as follows:

Given: $$F(x,y,z) = 0$$ Goal: Show $\frac{\partial z}{\partial y}|_x \frac{\partial y}{\partial x}|_z \frac{\partial x}{\partial z} |_y = -1$

Here is my work so far:

(1) Differentiate with respect to y, I get:

$0 + F_2 + F_3 \frac{\partial z}{\partial y} = 0$

So $ F_3 \frac{\partial z}{\partial y} = - F_2$

(2) Differentiate with respect to x, I get:

$F_1 + F_2 \frac{\partial y}{\partial x} + 0 = 0$

So $ F_2 \frac{\partial y}{\partial x} = - F_1$

(3) Differentiate with respect to z, I get:

$F_1 \frac{\partial x}{\partial z} + 0 + F_3 = 0$

So $ F_1 \frac{\partial x}{\partial z} = - F_3$

(4) After some manipulations with the $F_i$, I get to the conclusion that $\frac{\partial z}{\partial y}* \frac{\partial y}{\partial x} * \frac{\partial x}{\partial z} = -1$, so when evaluated with x, z, y respectively, conclusion is still true

My questions are:

(1) Is my proof correct?

(2) For example, when I differentiate with respect to y, I "let" $F_1$ be 0 and find partials for other coordinates.

I had a hard time trying to explain to my friend on the reason(s) why I can do such "let be 0" thing. Although I think if I can't do that, then there is no way that I can reach the conclusion, but I somehow feel confused about the fact too. Since my book is doing it that way, my understanding is that I can do such "let be 0" thing based on the independece of x with respect to y, when I differentiate with respect to y. But is my thought ok?

Would someone please help me on this question? Thank you very much ^_^

Solution 1:

A more simple way is the total differential $$dF=\frac {\partial F}{\partial x} dx+\frac {\partial F}{\partial y} dy+\frac {\partial F}{\partial z} dz=0$$ If it says "evaluated at $x$" it means that $x$ is fixed and $dx=0$ $$\frac {\partial F}{\partial y} dy+\frac {\partial F}{\partial z} dz=0\Rightarrow \frac{dz}{dy}=-\frac{\partial F/\partial y}{\partial F/\partial z}$$ and similarly $$dy=0\Rightarrow\frac {\partial F}{\partial x} dx+\frac {\partial F}{\partial z} dz=0\Rightarrow \frac{dx}{dz}=-\frac{\partial F/\partial z}{\partial F/\partial x}$$ $$dz=0\Rightarrow\frac {\partial F}{\partial x} dx+\frac {\partial F}{\partial y} dy=0\Rightarrow \frac{dy}{dx}=-\frac{\partial F/\partial x}{\partial F/\partial y}$$ And it follows $$\frac{dz}{dy}\frac{dy}{dx}\frac{dx}{dz}=\bigg(-\frac{\partial F/\partial y}{\partial F/\partial z}\bigg)\bigg(-\frac{\partial F/\partial x}{\partial F/\partial y}\bigg)\bigg(-\frac{\partial F/\partial z}{\partial F/\partial x}\bigg)=-1$$