Answer
(a) $\nabla f = \left( {{f_x},{f_y}} \right) = \left( {{y^2},2xy} \right)$, ${\ \ \ }$ $\nabla {f_{{\bf{r}}\left( t \right)}} = \left( {{t^6},{t^5}} \right)$
${\bf{r}}'\left( t \right) = \left( {t,3{t^2}} \right)$
(b) At $t=1$, $\frac{d}{{dt}}f\left( {{\bf{r}}\left( 1 \right)} \right) = 4$.
At $t=-1$, $\frac{d}{{dt}}f\left( {{\bf{r}}\left( { - 1} \right)} \right) = - 4$.
Work Step by Step
(a) We are given $f\left( {x,y} \right) = x{y^2}$ and ${\bf{r}}\left( t \right) = \left( {\frac{1}{2}{t^2},{t^3}} \right)$. So,
$\nabla f = \left( {{f_x},{f_y}} \right) = \left( {{y^2},2xy} \right)$, ${\ \ \ }$ $\nabla {f_{{\bf{r}}\left( t \right)}} = \left( {{t^6},{t^5}} \right)$
${\bf{r}}'\left( t \right) = \left( {t,3{t^2}} \right)$
(b) Using the Chain Rule for Paths we have
$\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right) = \nabla {f_{{\bf{r}}\left( t \right)}}\cdot{\bf{r}}'\left( t \right)$
$\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right) = \left( {{t^6},{t^5}} \right)\cdot\left( {t,3{t^2}} \right) = {t^7} + 3{t^7} = 4{t^7}$
At $t=1$, $\frac{d}{{dt}}f\left( {{\bf{r}}\left( 1 \right)} \right) = 4$.
At $t=-1$, $\frac{d}{{dt}}f\left( {{\bf{r}}\left( { - 1} \right)} \right) = - 4$.
