How to find the inverse of $f(x)=\dfrac{-x|x|}{1+x^2}$? I can see that 
$f(x)= \dfrac{-(x^2)}{1+x^2}$ for $x \geq 0$
and $f(x)= \dfrac{(x^2)}{1+x^2}$ for $x<0$.
Help me proceed to find the inverse. 
 A: The function
$$f:\quad x\mapsto y:=-{x\>|x|\over 1+x^2}\qquad(-\infty<x<\infty)$$
is odd and monotonically decreasing, see the following figure.

When $x\geq0$ we have $$y=-{x^2\over 1+x^2}\ ,\tag{1}$$ so that $f$ maps the $x$-interval $[0,\infty[\ $ onto the $y$-interval $\ ]-1,0]$. Solving $(1)$ for $x$ gives $x^2=-y/(1+y)$, hence
$$x=\sqrt{{-y\over 1+y}}\qquad(-1<y\leq0)\ ,$$
which is saying that
$$f^{-1}(y)=\sqrt{{-y\over 1+y}}\qquad(-1<y\leq0)\ .\tag{2}$$
Since $f$ is odd its inverse $f^{-1}$ is odd as well. We therefore have
$$f^{-1}(y)=-f^{-1}(-y)=-\sqrt{{y\over 1-y}}\qquad(0\leq y<1)\ .\tag{3}$$
The two formulas $(2)$ and $(3)$ can be condensed to
$$f^{-1}(y)=-{\rm sgn}(y)\sqrt{{|y|\over 1-|y|}}\qquad(-1<y<1)\ .$$
A: Solving for $x$ 
$$
y=\frac{x^2}{1+x^2}
$$
we have:
$$
x^2=\frac{y}{1-y}
$$
and for $x<0$ this gives 
$$
x=-\sqrt{\frac{y}{1-y}}
$$
Do the same for the other case.
A: You can usted the fact that f(x) is a Odd function, so if $x>0$ then $y=f(x)=\frac{-x^2}{1+x^2}$, so $y+yx^2=-x^2$ and $x^2=\frac{-y}{1+y}$ finally $f(x)^{-1}=\sqrt{\frac{-x}{1+x}}$ for $x>0$, and  $f(x)^{-1}=\sqrt{\frac{x}{1-x}}$ for $x\leq0$
