Show that the function $f(x)=x/(x^2-1)$ is bijective.

First notice that $$x=0\iff y=0.$$

Then for $|x|<1\land x\ne0$, $$y=\frac x{x^2-1}\iff yx^2-x-y=0.$$

The discriminant is strictly positive so that there are two distinct real roots. And by Vieta, their product is $-1$, so that one lies in $(-1,1)$ and the other not. Hence the function is invertible.

By using the quadratic formula one has $$f^{-1}(x)=\frac{1-\sqrt{4x^2+1}}{2x}$$ for the given range of $x$.