Reading this question $e^{C/x }-1=D/(x + a) $, i found my self completely unable to do anything. This is much more hard for me than my easy exercises about Lambert $W$-function.
So I probably need more training and exercise. Anyways thinking a bit about it I guess that the problem can be generalized in: find the inverse function $M^{-1} (x)$ of $M (x)=g(x)e^{f(x) }=y$ such that
$$g(M^{-1}(x))e^{f(M^{-1}(x)) }=x$$
and the easier case is when $W^{-1}(x)=xe^{x }=y$ and the inverse is the Lambert $W$
$$W(x) e^{W(x)}=x$$
at this point I can't not see that there is a relation between the solution of the first and the lambert function, and is that $M$ have a weird property
$$g(M^{-1}(x))e^{f(M^{-1}(x)) }=W(x) e^{W(x)}$$
then is possible to conclude that such such $M$ must hold:
$$f\circ M=g \circ M=W$$
but this does not make any sense (it does? If yes then for $M$ must hold $(f^{-1}\circ g)\circ M=M$ for every $f$ and $g$), and at this point I'm lost.
Questions
$1$) How we can find a general solution when $f$ and $g$ are invertible?
$$g(x)e^{f(x) }=y$$
$2$) What I am missing in my observation?