A side comment to start with: the notation cos^-1 for the inverse cosine function (more explicitly, the inverse function of the restriction of cosine to [0,pi]) is widespread but misleading. Indeed, the standard convention for exponents when using trig functions (e.g., cos^2 x:=(cos x)^2 suggests that cos^(-1) x is (cos x)^(-1)=1/(cos x). Of course, it is not, but the notation is very misleading. The alternative (and commonly used) notation arccos x is much better.
Now for the derivative. This is a composite, so we will use the Chain Rule. We will need (x^3)'=3x^2 and (arccos x)'=-1/sqrt(1-x^2) (see calculus of inverse trig functions ).
Using the Chain Rule:
(arccos(x^3))'=-1/sqrt(1-(x^3)^2) \times (x^3)'=-(3x^2)/sqrt(1-x^6) .
