Consider the task of finding the inverse function of the function $y=f(x)=\frac{x}{1-x^2}$ on $(-1,1)$.
One way is to try to isolate $x$ as a function of $y$. Sometimes it's easier to do this by switching the names of the variables and isolating $y$ as a function of $x$. $y=f^{-1}(x)$ is now the inverse of our original function.
$$x=\frac{y}{1-y^2}$$
$$xy^2+y-x=0\tag{1}$$
If $x=0$ then $y=0$
If $x\neq 0$ then we find that
$$y = \frac{-1 \pm \sqrt{1+4x^2}}{2x}$$
are candidate solutions.
$y = \frac{-1 + \sqrt{1+4x^2}}{2x}$ is the correct one. We can tell because in the original function, positive $x$ maps to positive $f(x)$ and negative $x$ maps to negative $f(x), and this behavior must be present in the inverse function. It is for this solution but not for the other solution.
My question here is about the (incorrect) solution to the problem
$$y = \frac{-1 - \sqrt{1+4x^2}}{2x}\tag{2}$$
Why exactly is it present?
Here is my attempt at an idea of why it might be present.
The function $f(x)=\frac{x}{1-x^2}$ is actually defined everywhere except at $1$ and $-1$. The graph looks like
$f$" />Does $(2)$ have something to do with the inverse of the portion of the graph of $f$ with $|x|>1$.
Note that since $\sqrt{1+4x^2}>1$ for $x\neq0$, then $-1-\sqrt{1+4x^2}<2$ and so
$$y = \frac{-1 - \sqrt{1+4x^2}}{2x}<-\frac{1}{x} \text{ for } x>0$$
$$y = \frac{-1 - \sqrt{1+4x^2}}{2x}>-\frac{1}{x} \text{ for } x<0$$