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?  
 A: No it possible to find an inverse. The approach has already had been shown but this will be easier for you to understand. First you should replace $x$ with $y$ and vice versa.
$\frac{y}{1-y^{2}}=x$
Multiply both sides by ${1-y^2}$.
$$x(1-y^2)=y$$
$$x-x{y^2}-y=0$$
$$(-x){y^2}+(-1){y}+(x)=0$$
Now compare this to $ay^2+by+c=0$. Taking the quadratic formula you can solve for $y$ and get two inverse formulas. 
So the inverse is split into...
$$y=\frac{-b+\sqrt{b^2-4ac}}{2a}$$ or $$y=\frac{-b-\sqrt{b^2-4ac}}{2a}$$.
Remeber depending on which inverse function you want to use, pay attention to the range and domain of the original function and its inverse. A graphing calculator like desmos can help.
A: You want to solve, with respect to $x$, the equation
$$
\frac{x}{1-x^2}=y
$$
with the condition that $-1<x<1$. If $y=0$, then $x=0$, so we can assume $y\ne0$. Thus the equation becomes
$$
x^2+\frac{1}{y}x-1=0
$$
Note that this equation has one negative and one positive solution. From the fact that $1-x^2>0$, we need to choose the positive solution for $y>0$ and the negative one for $y<0$. So we have
$$
x=\begin{cases}
\dfrac{-1/y-\sqrt{1/y^2+4}}{2} & \text{if $y<0$} \\[6px]
0 & \text{if $y=0$} \\[6px]
\dfrac{-1/y+\sqrt{1/y^2+4}}{2} & \text{if $y>0$}
\end{cases}
$$
Here's a diagram showing that the inverse function is correct:

If you follow the blue branch from $-\infty$ to $0$ and the red branch from $0$ to $\infty$, you get exactly the reflection around the bisector of the first and third quadrant of the graph of $f(x)=x/(1-x^2)$ restricted to $(-1,1)$.
Note that for $y>0$ we have
$$
0<\dfrac{-1/y+\sqrt{1/y^2+4}}{2}<1
$$
because, setting $t=1/y$ this is equivalent to
$$
\begin{cases}
\sqrt{t^2+4}>t\\
\sqrt{t^2+4}<2+t
\end{cases}
$$
The first inequality is obviously true; squaring the second one, we get
$$
t^2+4<4+4t+t^2
$$
which is true as well. Similarly, you can prove that, for $y<0$
$$
-1<\dfrac{-1/y-\sqrt{1/y^2+4}}{2}<0
$$
A: 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 $$x=\begin {cases} \frac {-1}{2y} + \frac {\sqrt {1+4y^2}}{2y} & y \neq 0\\0&y=0\end {cases}$$ The resulting curve is a nice reflection in $x=y$


A: Inverse function is 
$$ x = y/(1-y^2)$$
solving for y(x) we can choose any one of the two valid inverse functions
$$ y(x)= -\frac{1}{2x}\pm \sqrt { 1+\frac{1}{(2x)^2} }$$
