Answer
$\langle 2\sin t\cos t, 3\sin^2t \cos t\rangle $
Work Step by Step
Since $ r(t)=\langle t^2, t^3\rangle, \quad g(t)=\sin t $, then we have
$$ \frac{d}{dt}(r(g(t)))=r'(g(t)) g'(t)=\cos t\langle 2\sin t, 3\sin^2t\rangle=\langle 2\sin t\cos t, 3\sin^2t \cos t\rangle.$$
