How to find this function, and what method to use? The function is $f(x-\frac{1}{x})= x^3-\frac{1}{x^3}$ and they are asking us to find out what $f(-x)$ is? 

A general method for solving equations of the type:
$$f(g(x))=z(x)$$
Let:
$$u=x-\frac{1}{x}$$
Here we let $u$ be equal to $g(x)$.
$$ux=x^2-1$$
$$x^2-ux-1=0$$
Using the quadratic formula we have:
$$x=\frac{u \pm \sqrt{u^2+4}}{2}$$
Note in a way we found an inverse for $x-\frac{1}{x}$ even though it does not have an inverse function. Now we have:
$$f(u)=(\frac{u \pm \sqrt{u^2+4}}{2})^3-\frac{1}{(\frac{u \pm \sqrt{u^2+4}}{2})^3}$$
And:
$$f(x)=(\frac{x \pm \sqrt{x^2+4}}{2})^3-\frac{1}{(\frac{x \pm \sqrt{x^2+4}}{2})^3}$$
Which can be simplified down to get the result $f(x)=x^3+3x$.
$$f(-x)=(\frac{-x \pm \sqrt{x^2+4}}{2})^3-\frac{1}{(\frac{-x \pm \sqrt{x^2+4}}{2})^3}$$
Again this can be simplified to get your desired result $f(-x)=-x^3-3x$.
