<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] As you have observed, let:
$$h(a) = a(a+\frac{2}{x})$$ 
So solving the equation 
$$a = x - \frac{1}{x} $$
gives $$x = \frac{a \pm \sqrt{a^2 + 4}}{2}$$
Thus, we have $$h(a) = a(a + \frac{4}{a + \sqrt{a^2 + 4}})$$
(After we eliminate the minus case)
Which gives us the function that you are looking for. 
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