is it -3sin^3(3x) or did i miss something?
Yes you need to use the chain rule. Twice.
let
$\displaystyle f(u)=u^3$
$\displaystyle u=cos(v)$
and
$\displaystyle v=3x$
Then by the chain rule you get
$\displaystyle \frac{df}{dx}=\frac{df}{du}\frac{du}{dv}\frac{dv}{ dx}$
$\displaystyle 3u^2(-\sin(v))(3)=-9\cos^2(3x)\sin(x)$