Answer
$$ L(t)= \langle 0, \frac{\sqrt 2}{2} \rangle +t \langle-2, - \frac{3\sqrt 2}{2} \rangle .$$
Work Step by Step
Since $ r (t)=\langle \cos 2t,\sin 3t \rangle, $ then $ r' (t)=\langle -2\sin 2t,3\cos 3t \rangle $
then, the parametrization of the tangent line at $ t=\frac{\pi}{4}$ is given by
$$ L(t)=r (\frac{\pi}{4})+tr'(\frac{\pi}{4})=\langle 0, \frac{\sqrt 2}{2} \rangle +t \langle-2, - \frac{3\sqrt 2}{2} \rangle .$$
