How can I find the inverse of $y = \frac{x}{1-x^2}$?

Question

In example $1.5$ of Cracking the GRE Subject Test, the authors make the following calculation in one step with no additional commentary:

we interchange $x$ and $y$ and solve for $y$: \begin{align} \vdots\\ xy^2 + y-x &= 0\\ y &= \frac{-1 \pm \sqrt{1+4x^2}}{2x} \end{align}

What technique allows such a breezy solution?

Answer 1

Using the quadratic formula for $ay^2 + by +c = 0$

$$y = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

Here $a = x$, $b = 1$, $c = -1$

Answer 2

$$y = \frac{x}{1-x^2}$$

Replace $x$ with $y$

$$x = \frac{y}{1-y^2}$$

and solve $y,x=\frac{y}{1-y^2}$

$$-\frac{1\pm\sqrt{4x^2+1}}{2x}$$