Find $h\colon\Bbb R\setminus\{0\}\to \Bbb R$ with $h(x - \frac{1}{x})= x^2 - \frac{1}{x^2}$ for all $x\ne0$. 
Find $h\colon\Bbb R\setminus\{0\}\to \Bbb R$ with $h(x - \frac{1}{x})= x^2 - \frac{1}{x^2}$ for all $x\ne0$.

I saw instantly that $$h(x - \frac{1}{x})= x^2 - \frac{1}{x^2} = \left(  x -\frac{1}{x} \right)\left( x + \frac{1}{x} \right),$$ but I don't know how to proceed. I tried something like $$\left(x -\frac{1}{x} \right) = a$$ and $$\left( x + \frac{1}{x} \right) = a  + \frac{2}{x},$$ trying to apply $h(a)$ but it doesn't seem to work.
Any hints?

instead of writing x+1/x as a+2/x,
$ \sqrt{(x-1/x)^2+4} = \sqrt{a^2+4} = |x+1/x| = x+1/x$ for x>0 and $-(x+1/x)$ for x<0$
$h(a)=a \sqrt{a^2+4}$ for x>0 and $h(a)= -a \sqrt{a^2+4}$ for x<0, where a=x-1/x. 
Try simplifying this into a function in terms of x, considering cases for x>0 and x<0.
