Answer
$\frac{d^2y}{dt^2}=e^{-2t}(-5\sin3t-12\cos 3t).$
Work Step by Step
Since we have
$$ y= e^{-2t} \sin 3t $$
then the first derivative $\frac{dy}{dt}$, using the product rule, is given by
\begin{align*}
\frac{dy}{dt}&= (e^{-2t})' \sin3t+e^{-2t} (\sin 3t)'=-2e^{-2t} \sin3t+3e^{-2t} \cos 3t\\&=e^{-2t}( -2\sin 3t+ 3\cos 3t).\end{align*}
The second derivative $\frac{d^2y}{dt^2}$ is given by
\begin{align*}\frac{d^2y}{dt^2}&=(e^{-2t})'( -2\sin 3t+3\cos 3t)+e^{-2t}( -2\sin 3t+3\cos 3t)' \\
&=-2e^{-2t}( -2\sin 3t+3\cos 3t)+e^{-2t}(-6\cos 3t-9\sin3t)\\
&=e^{-2t}(-5\sin3t-12\cos 3t).
\end{align*}
