I am aware that inverting polynomials is hard and that probably there isn't a way to invert the function in general.
The question is whether this function can be inverted on a certain interval, for $x, \omega > 0$.
Particularly, I claim that the function is indeed invertible for $x \in [0, 1/\omega]$. This is because if I differentiate w.r.t $x$ and set to zero, I get two roots, the positive of which is $1/\omega$.
Question is whether the function can be inverted in closed form — i.e. in terms of elementary functions– over this interval, giving $f^{-1}(y) : [0, \frac{1}{\omega \left(1 + 1/\omega \right)^{1+\omega}}] \to [0, 1/\omega]$.
Best Answer
Let $$z=f(x)=\frac{x}{(1+x)^{1+\omega}}$$ and $$x=g(z)\equiv f^{-1}(z)=\sum^\infty_{n=1}g_n\frac{z^n}{n!}$$
Easily, by Lagrange inversion theorem about $z=f(a=0)$, $$g_n=(n+n\omega)^{\underline{n-1}}$$
Therefore, $$f^{-1}(z)=\sum^\infty_{n=1}\frac{(n+n\omega)^{\underline{n-1}} }{(n-1)!}\frac{z^n}n=\sum^\infty_{n=1}\binom{n+n\omega}{n-1}\frac{z^n}n$$
I doubt if you can get anything neater than this.