Chain rule for multiple variables?

Question

What I've tried so far:

$$F(x,y,z(x,y)) = 0$$ $$\implies \frac{\partial F}{\partial x} = 0$$

By the chain rule:

$$\frac{\partial F}{\partial x} = \frac{\partial F}{\partial z}\frac{\partial z}{\partial x} + \frac{\partial F}{\partial y}\frac{\partial z}{\partial y} + \frac{\partial F}{\partial x}\frac{\partial x}{\partial x} = 0$$ $$= \frac{\partial F}{\partial z}\frac{\partial z}{\partial x} + \frac{\partial F}{\partial y}\frac{\partial z}{\partial y} + \frac{\partial F}{\partial x} = 0$$

We know that $$\frac{\partial F}{\partial x} = 0$$ therefore

$$= \frac{\partial F}{\partial z}\frac{\partial z}{\partial x} + \frac{\partial F}{\partial y}\frac{\partial z}{\partial y} = 0$$

How do I proceed from here?

Answer 1

You have

$$0=\frac{\partial 0}{\partial x}= \frac{\partial F(x,y,z(x,y))}{\partial x}=F_x+F_z z_x\implies z_x=-\frac{F_x}{F_z}.$$

In a completely similar way

$$0=\frac{\partial 0}{\partial y}= \frac{\partial F(x,y,z(x,y))}{\partial y}=F_x+F_z z_y\implies z_y=-\frac{F_y}{F_z}.$$

Answer 2

Note that the you can always fix $y$ and change only $x$, then you get the same setting as in the theorem. Of course, you can also fix $x$ and let $y$ vary. In fact, also when you have $f:\mathbb{R}^n\to\mathbb{R}$, you can always fix all the variables except for two, and use the theorem.