Answer
$$ y''=2\sin t+4t \cos t-t^2\sin t,$$
$$ y'''=6\cos t-6t\sin t-t^2\cos t.$$
Work Step by Step
Since $ y=t^2\sin t $, then we have $ y'=2t \sin t+t^2\cos t=t(2\sin t+t\cos t)$ and by the product rule, we get
$$ y''= 2\sin t+t\cos t+t(3\cos t-t\sin t)=2\sin t+4t \cos t-t^2\sin t,$$
and again by using the product rule, we have
$$ y'''=2\cos t+4 \cos t-4t\sin t-2t\sin t-t^2\cos t=6\cos t-6t\sin t-t^2\cos t.$$