How to find the inverse function of the function $f$, defined as $f(x)=\frac{x}{1-x^2}$ for $-1 \lt x \lt 1$ I want to find the inverse function of the function $f$, defined as $f(x)=\frac{x}{1-x^2}$ for $-1 \lt x \lt 1$.
Letting $y=f(x)$, we first note that $x=0 \iff y=0$. Next, we consider the case when $y \neq 0$:

$y=\frac{x}{1-x^2} \implies-x^2y+y-x=0 \implies x^2+\frac{x}{y}-1=0 \implies \left(x+\frac{1}{2y}\right)^2-\left(1+\frac{1}{4y^2} \right)=0$

, which can be simplified to:

$\left|x+\frac{1}{2y} \right|=\sqrt{1+\frac{1}{4y^2}} \quad (*)$

If $x+\frac{1}{2y}\gt 0$, then we have: $g_1(y)=x=\sqrt{1+\frac{1}{4y^2}}-\frac{1}{2y}$
If $x+\frac{1}{2y} \lt 0$, then we have: $g_2(y)=x=(-1)\cdot\sqrt{1+\frac{1}{4y^2}}-\frac{1}{2y}$
In its current state, the collection of $(y,x)$ that satisfies $(*)$ does not represent a function...because for a single $y$, there is more than one $x$ that satisfies $(*)$.
Without making reference to graphing software, how does one determine the appropriate inverse for $f$?
Is there a standard procedure? One might argue that I should just try $g_1 \circ f$, $f \circ g_1$, $g_2 \circ f$, and $f \circ g_2$ and see which yield the $I$...but it seems to me like this doesn't necessarily exhaust all possibilities. For example, what if the true inverse is chopped up and pasted together versions of $g_1$ and $g_2$, so as to avoid the doubled values. i.e. let $g_3(x) = \begin{cases}g_1(x) &\text{ if } x \in (a,b)\\ g_2(x) &\text{ if }x \notin (a,b)\end{cases}$

I will retain the notation found in the original question...starting from:
$$\left|x+\frac{1}{2y} \right|=\sqrt{1+\frac{1}{4y^2}}$$
This expression means that for any $y \in \mathbb R \setminus \{0\}$:

$x=\sqrt{\left(1+\frac{1}{4y^2}\right)}-\frac{1}{2y} \ \lor\ x=(-1)\cdot\sqrt{\left(1+\frac{1}{4y^2}\right)}-\frac{1}{2y} \quad (*)$

First suppose that $y \lt 0$. Assume that  $x=\sqrt{\left(1+\frac{1}{4y^2}\right)}-\frac{1}{2y}$. If $y \lt 0$, then $-\frac{1}{2y} \gt 0$. Given that $\sqrt{\left(1+\frac{1}{4y^2}\right)} \gt 1$, we must have that $x \gt 1$.
Next, suppose that $y \gt 0$. Assume that $x=(-1)\cdot\sqrt{\left(1+\frac{1}{4y^2}\right)}-\frac{1}{2y}$. If $y \gt 0$, then $-\frac{1}{2y} \lt 0$. Given that $(-1)\cdot\sqrt{\left(1+\frac{1}{4y^2}\right)} \lt -1$, we must have that $x \lt -1$.
Now, our original function $f$ is defined on the open interval $(-1,1)$. This means that the inverse function $f^{-1}$ must necessarily have an image of $(-1,1)$.
In order for $(*)$ to be upheld, we must then have the following:
(1) If $y \lt 0$, then $x=(-1)\cdot\sqrt{\left(1+\frac{1}{4y^2}\right)}-\frac{1}{2y}$
(2) If $y \gt 0$, then $x=\sqrt{\left(1+\frac{1}{4y^2}\right)}-\frac{1}{2y}$
Letting $x = f^{-1}(y)$, we have the following description of $f^{-1}$:
$$f^{-1}(y)=\begin{cases}(-1)\cdot\sqrt{\left(1+\frac{1}{4y^2}\right)}-\frac{1}{2y}&\text { if } y\lt 0 \\0 &\text { if } y= 0\\\sqrt{\left(1+\frac{1}{4y^2}\right)}-\frac{1}{2y}&\text { if } y\gt 0 \end{cases}$$
