Answer
$$ y'=
-\sin( \theta)\sin(\cos \theta) \sin(\cos(\cos \theta)).$$
Work Step by Step
Recall that $(\cos x)'=-\sin x$.
Since $ y=\cos(\cos(\cos \theta ))$, then by the chain rule, the derivative $ y'$ is given by
$$ y'= -\sin(\cos(\cos \theta))(-\sin(\cos \theta))(-\sin \theta)\\=
-\sin( \theta)\sin(\cos \theta) \sin(\cos(\cos \theta)).$$
