Answer
$$\frac{d}{dt}(r(g(t)))=e^t\langle 2e^t, -1\rangle .$$
Work Step by Step
Since $ r(t)=\langle t^2, 1-t\rangle, \quad g(t)=e^t $, then we have $$\frac{d}{dt}(r(g(t)))=\frac{d}{dt}\langle e^{2t}, 1-e^t\rangle=\langle 2e^{2t}, -e^t\rangle \\=e^t\langle 2e^t, -1\rangle=r'(g(t)) g'(t).$$
