Answer
$\dfrac{d r}{d \tau}=\lt -3 \pi \sin (\pi \tau), 3 \pi \cos (\pi \tau) \gt$
Work Step by Step
Here, we have: $r(t)=\lt 3 \cos t, 3 \sin t\gt$ and $t=\pi \tau$
Apply Chain rule: $\dfrac{d r}{d \tau}=\dfrac{d r}{d t} \times \dfrac{d t}{d \tau}$
or, $\dfrac{d r}{d \tau}=\lt -3 \sin t, 3 \cos t\gt \times \pi=\lt -3 \pi \sin t, 3 \pi \cos t\gt$
By plugging $t=\pi \tau$ in the above equation, we get:
$\dfrac{d r}{d \tau}=\lt -3 \pi \sin (\pi \tau), 3 \pi \cos (\pi \tau) \gt$
