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?
First things first: Do not use the letter $x$ as independent variable of both $f$ and $f^{-1}$ as long as you study $f$ and $f^{-1}$ at the same time! Write $f^{-1}(y)=\ldots$ instead! This recommendation has nothing to do with the definitions of functions and inverses; it is just about letters. (Unfortunately there is no clear cut notion of "variable" in analysis; but this is another matter.)
The function $f$ is obviously defined for $-1<x<1$. It is the composition $f=h\circ g$ of the functions $$g:\quad x\mapsto u={x\over\sqrt{1-x^2}}\qquad{\rm and}\qquad h:\quad u\mapsto y:=e^{-\pi u}\ .\tag{1}$$ The function $g$ can be viewed as $g=\tan\circ\arcsin$, hence is strictly increasing, and maps $\>]{-1},1[\>$ bijectively to ${\mathbb R}$. The function $h$ is strictly decreasing, and maps ${\mathbb R}$ bijectively to ${\mathbb R}_{>0}$. It follows that $f=h\circ g:\ ]{-1},1[\>\to{\mathbb R}_{>0}$ is a decreasing bijective map, and has a well defined inverse $f^{-1}\!:\ {\mathbb R}_{>0}\to\>]{-1},1[\>$. No multivaluedness whatsoever will arise.
From $u^2(1-x^2)=x^2$ we obtain $x^2={u^2\over1+u^2}$, hence $$x=\pm{u\over\sqrt{1+u^2}}\ .$$ At this point we can definitively resolve the $\pm$-ambiguity by inspection of $(1)$: The variables $x$ and $u$ have the same signs at all times, since $\sqrt{1-x^2}$ is $\geq0$ by definition. It follows that $$x={u\over\sqrt{1+u^2}}\ .\tag{2}$$ On the other hand, from $y=e^{-\pi u}$ we immediately obtain $$u=-{1\over\pi}\log y\ .\tag{3}$$ Coupling $(2)$ and $(3)$ together we get $$x=f^{-1}(y)={-{1\over\pi}\log y\over\sqrt{1+({1\over\pi}\log y)^2}}={-\log y\over\sqrt{\pi^2+(\log y)^2}}\qquad(y>0)\ ,$$ without any cases and ambiguities.