Answer
$$ f'(x)
=(\cos^2x)^{\cos x}(-\sin x (\ln \cos^2x)+-2 \sin x).$$
Work Step by Step
Since $ f(x)=(\cos^2x)^{\cos x}$, applying $\ln $ on both sides, we get
$$\ln f(x)=\ln (\cos^2x)^{\cos x}=\cos x \ln (\cos^2x) $$
Hence the derivative, using the product rule, is given by
$$ f'/f =-\sin x (\ln \cos^2x)+\frac{\cos x}{\cos^2x}(\cos^2x)'\\
=-\sin x (\ln \cos^2x)+\frac{1}{\cos x} (-2\cos x\sin x)\\ =-\sin x (\ln \cos^2x)+-2\sin x $$
Then, we have
$$ f'(x)=f(x)(-\sin x (\ln \cos^2x)+-2\sin x)\\
=(\cos^2x)^{\cos x}(-\sin x (\ln \cos^2x)+-2 \sin x).$$
