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?

First you must solve 
$x=y-\frac{1}{y}$ 
$y^2-xy-1=0$ 
So
$y=\frac{x+\sqrt{x^2+4}}{2}$
Then
$h(x)=h(y-\frac{1}{y})=y^2-\frac{1}{y^2}=(y-\frac{1}{y})(y+\frac{1}{y})=$
$=(\frac{x+\sqrt{x^2+4}}{2}-\frac{2}{x+\sqrt{x^2+4}})(\frac{x+\sqrt{x^2+4}}{2}+\frac{2}{x+\sqrt{x^2+4}})$
