I've been trying to find the inverse of $$f(x) = e^{-\left(\displaystyle \frac{x}{\sqrt{1-x^2}}\right) \displaystyle \pi }$$
Here are my steps $$ \begin{align} x & = e^{-\left(\displaystyle \frac{y}{\sqrt{1-y^2}}\right) \displaystyle \pi }\\\\ \displaystyle \ln(x) & = \left(\displaystyle -\frac{y\pi}{\sqrt{1-y^2}}\right) \\\\ \displaystyle \left(\ln(x)\right)^2 & = \left(\displaystyle -\frac{y\pi}{\sqrt{1-y^2}}\right)^2 \\\\ \displaystyle \ln^2(x) & = \displaystyle \frac{y^2\pi^2}{1-y^2} \\\\ \displaystyle \ln^2(x) - \ln^2(x) y^2 & = y^2\pi^2 \\\\ \displaystyle \ln^2(x) & = y^2\left[\pi^2 + \ln^2(x)\right] \\\\ \displaystyle y^2 & = \frac{ \ln^2(x)} {\pi^2 + \ln^2(x)} \\\\ \displaystyle y & = \pm \sqrt{ \frac{\ln^2(x)} {\pi^2 + \ln^2(x)} }\\\\ \displaystyle f^{-1}(x) & = \pm \sqrt{ \frac{\ln^2(x)} {\pi^2 + \ln^2(x)} }\\\\ \end{align} $$
Now by looking into the graph that I've been made on desmos

By looking at the orange and purple line, I can conlude the result $$ f^{-1}(x) = \begin{cases} \hfill \sqrt{ \frac{\ln^2(x)} {\pi^2 + \ln^2(x)} } \hfill & {\text{if}}\ 0<x\leq1 \\ \hfill -\sqrt{ \frac{\ln^2(x)} {\pi^2 + \ln^2(x)} } \hfill & {\text{if}}\ x>1 \\ \hfill \text{undefined}\ \hfill & \text{if else} \\ \end{cases} $$
Now, without graphical approach, how do I find the piece-wise result?