Find $h\colon\Bbb R\setminus\{0\}\to \Bbb R$ with $h(x - \frac{1}{x})= x^2 - \frac{1}{x^2}$ for all $x\ne0$.

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.

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.