I try to find the inverse function of $\sqrt[1-x]{x}$ for the interval $(0, 1)$. Given the Lambert W function, I was able to derive a possible candidate $\frac{W\left(y\ \cdot\ln\left(y\right)\right)}{\ln\left(y\right)}$ working already perfectly for values in the interval $(0, 1/e)$. Unfortually, from this point onwards the function just returns values nearby 1 for all inputs.
Is there any alternative inverse function (even piecewise or just as an approximation) for all the values in the interval $(0, 1)$?
$$y(x)=\sqrt[1-x]{x}$$ $$y(x)=x^{\frac{1}{1-x}} $$ You are right. The inverse function is (formaly) :
$$x(y)=\frac{W\left(y\ \cdot\ln\left(y\right)\right)}{\ln\left(y\right)} \tag 1$$ But don't forget that the Lambert $W(X)$ function is multivalued for $-\frac{1}{e} < X < 0$ .
The two real branches are :
The common branch denoted with subscribt $0$ : $$W_0(X) \quad \text{for}\quad -\frac{1}{e}<X<\infty$$ and the second branch with subscript $-1$ : $$W_{-1}(X) \quad \text{for}\quad -\frac{1}{e} < X < 0$$
As a consequence in order to get the whole inverse function :
$$x(y)=\frac{W_0\left(y\ \cdot\ln\left(y\right)\right)}{\ln\left(y\right)}\qquad 0 < y < \frac{1}{e}$$
$$x(y)=\frac{W_{-1}\left(y\ \cdot\ln\left(y\right)\right)}{\ln\left(y\right)}\qquad \frac{1}{e} < y < 1$$
Note that $X>0$ corresponding to $y>1$ is not used in the present case : $x(y > 1)=1.$
If you use WolframAlpha the syntax is LambertW(0,X) and LambertW(-1,X) .