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. 

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 $x\leq0$
