Answer
\begin{align*}\frac{d^2y}{dt^2}
&=2e^t\cos t.
\end{align*}
Work Step by Step
Recall the product rule: $(uv)'=u'v+uv'$
Recall that $(e^x)'=e^x$
Recall that $(\sin x)'=\cos x$.
Since we have
$$ y= e^t \sin t $$
then the first derivative $\frac{dy}{dt}$, using the product rule, is given by
\begin{align*}
\frac{dy}{dt}&= (e^t)' \sin t+e^t (\sin t)'=e^t \sin t+e^t \cos t\\&=e^t( \sin t+ \cos t).\end{align*}
The second derivative $\frac{d^2y}{dt^2}$ is given by
\begin{align*}\frac{d^2y}{dt^2}&=(e^t)'( \sin t+ \cos t)+e^t( \sin t+ \cos t)' \\
&=e^t( \sin t+ \cos t)+e^t( \cos t- \sin t)\\
&=2e^t\cos t.
\end{align*}
