Sufficient conditions using implicit function theorem

Question

Let $f:\mathbb{R^3} \to \mathbb{R}.$

Let $ (x_0,y_0,z_0)$ be a solution of the equation $f(x-y,y-z,z-x)=0$.

Find sufficient conditions such that extracting $z$ with respect to $x,y$ is possible.

I study about implicit function theorem and I find out this theorem but the equation $f(x-y,y-z,z-x)=0$ gets me a little bit confused.

My attempt:

For applying implicit function theorem in $(a,b,c)$ I have to demand:

1.$f(a,b,c)=0$.

2.There is a neigborhood $U$ of $(a,b,c)$ such that $f$ is $C^1$ in $U$ and

3.$\frac{\partial f}{\partial z}(a,b,c)\neq 0$

Denote $a:=x_0-y_0,b:=y_0-z_0,c:=z_0-x_0 \implies f(a,b,c)=0$.

$f\in C^1$.

I have an issue finding the derivative of $f$.

Is my solution correct?

Answer 1

You can define $F(x,y,z) = f(x-y,y-z,z-x)$ and $\phi(x,y,z)= (x-y, y-z,z-x)$. You have $F(x,y,z) = (f \circ \phi)(x,y,z)$. From there, you just have to apply the chain rule:

$$\frac{\partial F}{\partial z}(x,y,z) = -\frac{\partial f}{\partial y}(x-y,y-z,z-x) + \frac{\partial f}{\partial z}(x-y,y-z,z-x).$$ Finally, the condition you're looking for is

$$\frac{\partial f}{\partial y}(x_0-y_0,y_0-z_0,z_0-x_0) \neq \frac{\partial f}{\partial z}(x_0-y_,y_0-z_0,z_0-x_0)$$ applying implicit function theorem.

Answer 2

It is not $\frac{\partial f}{\partial z}(a,b,c)$ which must be non-zero. It is $\frac{\partial g}{\partial z}(x_0,y_0,z_0),$ where $$g(x,y,z)=f(x-y,y-z,z-x).$$ To compute $\frac{\partial g}{\partial z},$ use the chain rule: $$\frac{\partial g}{\partial z}(x,y,z)=(\partial_3f-\partial_2f)(x-y,y-z,z-x).$$