Answer
$$\left\langle {2t{e^t}, - 2{e^t},0} \right\rangle $$
Work Step by Step
$$\eqalign{
& {\text{Let }}{\bf{v}}\left( t \right) = \left\langle {{t^2}, - 2t,1} \right\rangle \cr
& {\text{Calculate }}\frac{d}{{dt}}\left[ {{\bf{v}}\left( {{e^t}} \right)} \right].{\text{ Use }}\frac{d}{{dt}}\left[ {{\bf{v}}\left( {f\left( t \right)} \right)} \right] = {\bf{v}}'\left( {f\left( t \right)} \right)f'\left( t \right) \cr
& {\bf{v}}'\left( t \right) = \left\langle {2t, - 2,0} \right\rangle \cr
& f'\left( t \right) = \frac{d}{{dt}}\left[ {{e^t}} \right] = {e^t} \cr
& {\text{Thus}}{\text{,}} \cr
& \frac{d}{{dt}}\left[ {{\bf{v}}\left( {{e^t}} \right)} \right] = \left\langle {2t, - 2,0} \right\rangle \left( {{e^t}} \right) \cr
& {\text{Multiplying}} \cr
& \frac{d}{{dt}}\left[ {{\bf{v}}\left( {{e^t}} \right)} \right] = \left\langle {2t{e^t}, - 2{e^t},0} \right\rangle \cr} $$
