Inverse of $f(x)=\dfrac{x}{4-x^2}$? I tried writing
$x=\dfrac{f^{-1}(x)}{4-[f^{-1}(x)]^2}$
but can't make $f^{-1}(x)$ subject of the formula.

$$y=f(x)=\frac{x}{4-x^2}$$
$$y(4-x^2)={x}$$
$$4y-yx^2={x}$$
$$yx^2+x-4y=0$$
Now use quadratic formula
$$x=\frac{-1\pm\sqrt{1+16y^2}}{2y}$$

$$f^{-1}(x)=\frac{-1\pm\sqrt{1+16x^2}}{2x}$$

