Answer
\begin{align*}\frac{d^3y}{d\theta}& =
e^{3 \theta} \sin \left(e^{\theta}\right)-3 e^{2 \theta} \cos \left(e^{\theta}\right)-e^{\theta} \sin \left(e^{\theta}\right).
\end{align*}
Work Step by Step
Recall the product rule: $(uv)'=u'v+uv'$
Recall that $(e^x)'=e^x$
Recall that $\cos(-x)=\cos(x)$.
Recall that $\sin(-x)=-\sin(x)$.
Since we have
$$ y= \cos (e^\theta )$$
then the first derivative $\frac{dy}{d\theta}$, using the chain rule, is given by
\begin{align*}
\frac{dy}{d\theta}&= -\sin(e^\theta)(e^\theta)'\\
&= -e^\theta \sin(e^\theta).
\end{align*}
The second derivative $\frac{d^2y}{d\theta}$, using the product rule, is given by
\begin{align*}\frac{d^2y}{d\theta^2}&=-(e^\theta)'\sin(e^\theta)-(e^\theta)(\sin(e^\theta))' \\
&=-e^\theta\sin(e^\theta)-e^{2\theta} \cos(e^\theta)\\
&=-e^\theta(\sin(e^\theta)+e^{\theta} \cos(e^\theta)).
\end{align*}
The third derivative $\frac{d^3y}{d\theta}$, using the product rule, is given by
\begin{align*}\frac{d^3y}{d\theta}&=-(e^\theta)' (\sin(e^\theta)+e^{\theta} \cos(e^\theta))-e^\theta(\sin(e^\theta)+e^{\theta} \cos(e^\theta))' \\
&=-e^\theta (\sin(e^\theta)+e^{\theta} \cos(e^\theta))-e^\theta(e^\theta\cos(e^\theta)+e^\theta\cos(e^\theta)-e^{2\theta} \sin(e^\theta)) \\
&=
e^{3 \theta} \sin \left(e^{\theta}\right)-3 e^{2 \theta} \cos \left(e^{\theta}\right)-e^{\theta} \sin \left(e^{\theta}\right).
\end{align*}
