Answer
$$\left\langle {0,3{t^2},6{t^5}} \right\rangle $$
Work Step by Step
$$\eqalign{
& {\text{Let }}{\bf{u}}\left( t \right) = \left\langle {1,t,{t^2}} \right\rangle \cr
& {\text{Calculate }}\frac{d}{{dt}}\left[ {{\bf{u}}\left( {{t^3}} \right)} \right].{\text{ Use }}\frac{d}{{dt}}\left[ {{\bf{u}}\left( {f\left( t \right)} \right)} \right] = {\bf{u}}'\left( {f\left( t \right)} \right)f'\left( t \right) \cr
& {\bf{u}}'\left( t \right) = \left\langle {0,1,2t} \right\rangle \cr
& {\bf{u}}'\left( {{t^3}} \right) = \left\langle {0,1,2{t^3}} \right\rangle \cr
& f'\left( t \right) = \frac{d}{{dt}}\left[ {{t^3}} \right] = 3{t^2} \cr
& {\text{Thus}}{\text{,}} \cr
& \frac{d}{{dt}}\left[ {{\bf{u}}\left( {{t^3}} \right)} \right] = \left\langle {0,1,2{t^3}} \right\rangle \left( {3{t^2}} \right) \cr
& {\text{Multiplying}} \cr
& \frac{d}{{dt}}\left[ {{\bf{u}}\left( {{t^3}} \right)} \right] = \left\langle {0,3{t^2},6{t^5}} \right\rangle \cr} $$
