Answer
$$\frac{dy}{dx}=-6\cos^2(\sin2x)\sin(\sin2x)\cos2x$$
Work Step by Step
$$\frac{dy}{dx}=(\cos^3(\sin2x))'=3\cos^2(\sin2x)\cdot(\cos(\sin2x))'=
3\cos^2(\sin2x)(-\sin(\sin2x)\cos(\sin2x)'=-3\cos^2(\sin2x)\sin(\sin2x)\cos2x(2x)'=
-3\cos^2(\sin2x)\sin(\sin2x)\cos2x\cdot2=-6\cos^2(\sin2x)\sin(\sin2x)\cos2x$$
