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: Observe that $f(x-\frac{1}{x})= x^3-\frac{1}{x^3}=(x-\frac{1}{x})^3+3\cdot x\cdot \frac{1}{x}\cdot(x-\frac{1}{x})=(x-\frac{1}{x})^3+3(x-\frac{1}{x})$
Hence, we can say that $$f(z)=z^3+3z$$ where $z$ is any real variable.
So we have that $$\color{red}{f(-x)}=\color{blue}{-x^3-3x}$$
A: A slightly different approach from the others:  
Note that every real number $x$ can be written in the form $a-\frac1a$ for some $a\ne0$. Then $-x=-\left(a-\frac1a\right)=(-a)-\frac1{-a}$, so $$f(-x)=(-a)^3-\frac1{(-a)^3}=-\left(a^3-\frac1{a^3}\right)=-f(x).$$  
If you  need an explicit expression for $f(-x)$, then either solve for $a$ and substitute, or use one of the methods described in the other answers.
A: 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 $f(-x)=-x^3-3x$.
A: write it as follows $$f\left( x-\frac { 1 }{ x }  \right) =x^{ 3 }-\frac { 1 }{ x^{ 3 } } ={ \left( x-\frac { 1 }{ x }  \right)  }^{ 3 }+3\left( x-\frac { 1 }{ x }  \right) $$ it means our function is $$f\left( x \right) ={ x }^{ 3 }+3x$$ 
now 

$$f\left( -x \right) =-{ x }^{ 3 }-3x$$

A: As per Parth Kohli's comment we can write $$f\left(x - \frac{1}{x}\right) = \left(x - \frac{1}{x}\right)^3 + 3\left(x - \frac{1}{x}\right).$$
That is, $f(x) = x^3 + 3x$ so $f(-x) = (-x)^3 + 3(-x) = -(x^3 + 3x)$. 
