Suppose $f(x)-f(y)=h(g(x)-g(y))$, then $h$ is linear?

Fix $c$. If

$$f(x)-f(y)=h(g(x)-g(y))$$

then

$$f(x)-f(y)=h(g_{2}(x)-g_{2}(y))$$

where

$$g_{2}(x)=g(x)-g(c)$$.

I think that completes your argument.

Neat question.