Answer
$\langle 16t\cos 2t^2, -24\sin2t^2\rangle $
Work Step by Step
Since $ r(t)=\langle 4\sin 2t, 6\cos 2t \rangle, \quad g(t)=t^2 $, then we have $$ \frac{d}{dt}(r(g(t)))=r'(g(t)) g'(t)=2t \langle 8\cos 2t^2, -12\sin2t^2\rangle= \langle 16t\cos 2t^2, -24\sin2t^2\rangle.$$
