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?

There is no function satisfying $h(x - \frac{1}{x})= x^2 - \frac{1}{x^2}$ for all $ x \ne 0$.
For any $a \ne 0$ the equation  $x - \frac 1x = a$ has two solutions $x_1, x_2$, which are related by $x_1 x_2 = -1$. Then
$$
 x_1^2 - \frac{1}{x_1^2} = h(a) =  x_2^2 - \frac{1}{x_2^2} = \frac{1}{x_1^2} - x_1^2 \\
\implies x_1^4 = 1 \implies x_1 = \pm 1 \\
\implies a= 0 \, ,
$$
a contradiction. 
More concretely, for $x=2$ you would get
$$
 h(\frac 32) = h(2 - \frac 12) = 4 - \frac 14 = \frac{15}{4}
$$
and for $x = -1/2$
$$
h(\frac 32) = h(-\frac 12 + 2) = \frac 14 - 4 = -\frac{15}{4}
$$
which is a clear contradiction.
