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