Answer
$$y'=\frac{-3 \sin \left(e^{\sqrt{\tan 3 x}}\right) e^{\sqrt{\tan 3 x}} \sec ^{2}(3 x)}{2 \sqrt{\tan 3 x}} $$
Work Step by Step
Given $$y=\cos(e^{\sqrt{\tan 3x}})$$
Then
\begin{align*}
y^{\prime} &=-\sin \left(e^{\sqrt{\tan 3 x}}\right) \cdot\left(e^{\sqrt{\tan 3 x}}\right)^{\prime}\\
&=-\sin \left(e^{\sqrt{\tan 3 x}}\right) e^{\sqrt{\tan 3 x}} \cdot \frac{1}{2}(\tan 3 x)^{-1 / 2} \cdot \sec ^{2}(3 x) \cdot 3 \\
&=\frac{-3 \sin \left(e^{\sqrt{\tan 3 x}}\right) e^{\sqrt{\tan 3 x}} \sec ^{2}(3 x)}{2 \sqrt{\tan 3 x}}
\end{align*}