$\phi(x)=f(x)/g(x)$. Then $\phi(x)=h(f(x))$ if and only if $g(x)=\psi(f(x))$ for some function $psi$?

Question

$f,g,\phi$ are 2D non-constant real functions. $x\in\mathbb R^2$

Given: $\phi(x)=\frac{f(x)}{g(x)}$.

Then is it possible to claim that $\phi(x)=h(f(x))$ (for some real function $h$), if and only if $g(x)=\psi(f(x))$ for some real function $\psi$?

The question sounds very simple. For example, let $\phi(x)=\frac{x_1+x_2}{(x_1+x_2)^2}$, then we simply have $h(y)=\frac{y}{y^2}$.

But if $\phi(x)=\frac{x_1+x_2}{x_1-x_2}$, then it seems like that we cannot find a $h$.

The "if" direction is easy. I have no clue for the "only if" direction. Not even sure if it is easily provable.

Answer 1

For all $x$ such that $f(x)\ne0$, $$g(x)=\frac{f(x)}{f(x)/g(x)}=\frac{f(x)}{h(f(x))}$$

On the other hand, consider two functions $f$, $g$ and a point $c$ such that $f(c)=0$ and $g(c)\ne0$. Then, $$v(x)=\begin{cases}g(x)&\text{if }x\ne c\\ \lvert g(x)\rvert+1&\text{if }x=c\end{cases}$$

satisfies $\frac{f(x)}{v(x)}=\frac{f(x)}{g(x)}$ for all $x$ in the domain of $\frac fg$.