How to make squaring method of finding an inverse function to be invertible?

Question

I've been trying to find an inverse of this function

$$f(x) = e^{-\left(\displaystyle \frac{x}{\sqrt{1-x^2}}\right) \displaystyle \pi }$$

These are the approaches

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First approach using squaring method

$$ \begin{align} y & = e^{-\left(\displaystyle \frac{x}{\sqrt{1-x^2}}\right) \displaystyle \pi }\\ \displaystyle \ln(y) & = \left(\displaystyle -\frac{x\pi}{\sqrt{1-x^2}}\right) \\ \displaystyle \left(\ln(y)\right)^2 & = \left(\displaystyle -\frac{x\pi}{\sqrt{1-x^2}}\right)^2 \\ \displaystyle \ln^2(y) & = \displaystyle \frac{x^2\pi^2}{1-x^2} \\ \displaystyle \ln^2(y) - \ln^2(y) x^2 & = x^2\pi^2 \\ \displaystyle \ln^2(y) & = x^2\left[\pi^2 + \ln^2(y)\right] \\ \displaystyle x^2 & = \frac{ \ln^2(y)} {\pi^2 + \ln^2(y)} \\ \displaystyle x & = \pm \sqrt{ \frac{\ln^2(y)} {\pi^2 + \ln^2(y)} }\\ \displaystyle \therefore\qquad f^{-1}(y)&={\pm\sqrt{\frac{\ln^2(y)}{\pi^2+\ln^2(y)}}}\\ \end{align} $$

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Second approach using function composition method

Source by Christian Blatter answer

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.

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Third approach by using Symbolab in here

                  

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Fourth apporach by using WolframAlpha in here

                                     

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Using squaring method the inverse is multivalued and has two value, therefore not invertible

Symbolab is also multivalued inverse, it's weird that it has four value, also not invertible

WolframAlpha is also multivalued inverse, but it only plot the negative value of it, it's weird, why even bother put $\pm$ sign in the first place?

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This is related to my question before. In that answer, the approach of using function decomposition method seems to be the only ones that truly give the solution that is invertible. However, it is tedious and not straightforward as squaring method.

So my question, is there any way to make the squaring method approach above invertible? Is there any mistake that I made that make it not invertible?

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Edit

After seen the Somos answer and Lee David Chung Lin comments, i think it may be better to put their answer in this question itself

\begin{array}{rl} \displaystyle f(x) = \;\; y&= e^{-\left(\displaystyle \frac{x}{\sqrt{1-x^2}}\right) \displaystyle \pi }\\\\ \displaystyle \ln(y) & = \left(\displaystyle -\frac{x\pi}{\sqrt{1-x^2}}\right) \qquad\qquad (*) \\\\ \displaystyle \left(\ln(y)\right)^2 & = \left(\displaystyle -\frac{x\pi}{\sqrt{1-x^2}}\right)^2 \\\\ \displaystyle \ln^2(y) & = \displaystyle \frac{x^2\pi^2}{1-x^2} \\\\ \displaystyle \ln^2(y) - \ln^2(y) x^2 & = x^2\pi^2 \\\\ \displaystyle \ln^2(y) & = x^2\left[\pi^2 + \ln^2(y)\right] \\\\ \displaystyle x^2 & = \displaystyle \frac{ \ln^2(y)} {\pi^2 + \ln^2(y)} \\\\ \displaystyle x & = \displaystyle \pm \sqrt{ \frac{\ln^2(y)} {\pi^2 + \ln^2(y)} }\\\\ \displaystyle x & = \displaystyle \frac{ \pm\sqrt{\ln^2(y)}} {\sqrt{\pi^2 + \ln^2(y)} }\\\\\\ \end{array}

$\text{Denumerator } \quad \rightarrow \sqrt{\pi^2 + \ln^2(y)} \;\; \text{ is always positive}\\ \text{Numerator } \qquad \rightarrow \pm\sqrt{\ln^2(y)} \qquad\,\text{ is related to } (*)$ \begin{array}{c} \qquad\qquad\qquad\qquad\qquad\quad\;\displaystyle \text{ if } x \text{ is positive, then } \ln(y) = \ln(f(x)) \text{is negative}\\ \qquad\qquad\qquad\qquad\qquad\quad\;\displaystyle \text{and}\\ \qquad\qquad\qquad\qquad\qquad\quad\;\,\displaystyle \text{ if } x \text{ is negative, then } \ln(y) = \ln(f(x)) \text{is positive }\\\\\\ \end{array}

$$\displaystyle \therefore\qquad f^{-1}(y) = x = \frac{- \ln(y)}{\sqrt{\pi^2 + \ln^2(y)}}$$

The equality must be persistent, such that $+ = +$ and $- = -$

Answer 1

Given the definition of $\,f(x),\,$ then let us define $$ h(x) := \ln(y) = \ln(f(x)) = -\Bigg(\frac{x}{\sqrt{1-x^2}}\Bigg)\pi. \tag1$$ Notice that if $\, x > 0\,$ then $\, h(x) < 0.\,$ In your last step you have, $$ x = f^{-1}(y) ={\pm\sqrt{\frac{\ln^2(y)}{\pi^2+\ln^2(y)}}} = \frac{\pm\sqrt{\ln^2(y)}}{\sqrt{\pi^2+\ln^2(y)}}. \tag2$$ The numerator is either $\,+\ln(y)\,$ or $\,-\ln(y),\,$ but it must be $\, -\ln(y).\,$ Reason is the denominator is positive and if $\,x = f^{-1}(y) >0\,$ then the numerator must be positive also since $\, h(x) = \ln(y) < 0 \,$ for $ x>0.$