What is exactly the inverse of the function $f(x)=\frac{x}{1-x^2}$? How do we find the inverse of the function $f:(-1,1)\to \Bbb R$ by $f(x)=\frac{x}{1-x^2}$?

The problem has been posted here and the answer below says "To show that $f$ has a continuous inverse, you just need to show that $f$ is an open map..." 

Is it impossible to find the inverse analytically?
For there to be an inverse, the original function has to pass the horizontal line test-there must not be two points on the curve with the same $y$ value.  Your function satisfies that.  A graph (below) is not a proof, but very suggestive.  In this case you can prove it by showing $y' \gt 0$, which shows the function is monotonic.  If you solve $y=\frac x{1-x^2}$ for $x$ you get $$y=\frac x{1-x^2}\\y-x^2y=x\\x^2y+x-y=0\\x=\frac {-1}{2y} \pm \frac {\sqrt {1+4y^2}}{2y}$$  You need to recognize that you can't divide by zero when $y=0$.  We resolve the sign ambiguity by noting that the signs of $x$ and $y$ are always the same.  This will be true with the plus sign so the inverse is the unique $$x=\begin {cases} \frac {-1}{2y} + \frac {\sqrt {1+4y^2}}{2y} &amp; y \neq 0\\0&amp;y=0\end {cases}$$ The resulting curve is a nice reflection in $x=y$
Inverse function is 
$$ x = y/(1-y^2)$$
solving for y(x) we can choose any one of the two valid inverse functions
$$ y(x)= -\frac{1}{2x}\pm \sqrt { 1+\frac{1}{(2x)^2} }$$