Answer
$$ L(s)= \langle 2,0, -\frac{1}{3}\rangle +s \langle-1,0,\frac{1}{2}\rangle .$$
Work Step by Step
Since we can write $ r (s)=\langle 4s^{-1},0, -\frac{8}{3}s^{-3}\rangle, $ then we have the derivative vector $ r' (s)=\langle -4s^{-2},0,8s^{-4} \rangle $
Then, the parametrization of the tangent line at $ s=2$ is given by
$$ L(s)=r (2)+sr'(2)=\langle 2,0, -\frac{1}{3}\rangle +s \langle-1,0,\frac{1}{2}\rangle .$$
