In general, to find the inverse of $f(x)$, one solves $x = f(y)$ for $y$. In your case that yields
$$
x = \frac{y}{y^2-1} \Leftrightarrow xy^2 -y - x = 0
$$
for which you can apply the quadratic formula
$$
y_\pm = \frac{-1 \pm \sqrt{1-4x^2}}{2x}
$$
So there are 2 inverse functions.
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Bumbble Comm
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All you have to do is to solve the equation $\frac x{x^2-1}=a$ and to find a solution in $(-1,1)$.
In general, to find the inverse of $f(x)$, one solves $x = f(y)$ for $y$. In your case that yields $$ x = \frac{y}{y^2-1} \Leftrightarrow xy^2 -y - x = 0 $$ for which you can apply the quadratic formula $$ y_\pm = \frac{-1 \pm \sqrt{1-4x^2}}{2x} $$ So there are 2 inverse functions.