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) #.
