Consider

$\displaystyle \frac{x}{x^2-1} = \alpha \implies f(x) = \alpha x^2 - x -\alpha = 0$ when $\displaystyle \alpha = 0, x= 0$ $\displaystyle \alpha \neq 0$ we solve for $\displaystyle x$ we have $\displaystyle x = \frac{1 \pm \sqrt{1 + 4 \alpha^2}}{2\alpha}$. Now we are not sure as to which root to select.

Consider $\displaystyle \alpha > 0$ figure 1 shows thw plot of the quadratic where we need the roots in (-1,1) so this requires $\displaystyle f(-1)> 0 , f(1)> 0$. but we have f(1) = -1 and f(-1)=1. $\displaystyle \therefore$ we choose $\displaystyle \beta = \frac{1 - \sqrt{1 + 4 \alpha^2}}{2\alpha}$ $\displaystyle \ni \beta \in (-1,1) , f(\beta) > 0$ and $\displaystyle f(\beta) = \alpha$. Similarly you can argue from figure 2 that for $\displaystyle \alpha < 0$ choose $\displaystyle \beta = \frac{1 + \sqrt{1 + 4 \alpha^2}}{2\alpha}$. Hence it is onto.