<s>[INST] 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?
 [/INST] Let
$$x=y-\frac1y$$
so that
$$y=\frac{x\pm\sqrt{x^2+4}}2.$$
Then
$$f(x)=f\left(y-\dfrac1y\right)=y^2-\frac1{y^2}=\left(\frac{x\pm\sqrt{x^2+4}}2\right)^2-\left(\frac2{x\pm\sqrt{x^2+4}}\right)^2\\
=\left(\frac{x\pm\sqrt{x^2+4}}2\right)^2-\left(\frac{x\mp\sqrt{x^2+4}}2\right)^2=\pm x\sqrt{x^2+4}
.$$
Note that, as raised by Martin, to define a function only one sign at a time is possible and we have to choose. Then the sign of $y$ is constant and the domain of $x$ is restricted.
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