Answer
$\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right){|_{t = 2}} = 3 + 4{{\rm{e}}^4}$
Work Step by Step
We have $f\left( {x,y} \right) = x + {{\rm{e}}^y}$ and ${\bf{r}}\left( t \right) = \left( {3t - 1,{t^2}} \right)$.
So,
$\nabla f = \left( {1,{{\rm{e}}^y}} \right)$, ${\ \ \ \ }$ $\nabla {f_{{\bf{r}}\left( t \right)}} = \left( {1,{{\rm{e}}^{{t^2}}}} \right)$
${\bf{r}}'\left( t \right) = \left( {3,2t} \right)$
At $t=2$, we get $\nabla {f_{{\bf{r}}\left( 2 \right)}} = \left( {1,{{\rm{e}}^4}} \right)$ and ${\bf{r}}'\left( 2 \right) = \left( {3,4} \right)$.
Using the Chain Rule for Paths (Theorem 2 of Section 15.5):
$\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right){|_{t = 2}} = \nabla {f_{{\bf{r}}\left( 2 \right)}}\cdot{\bf{r}}'\left( 2 \right) = \left( {1,{{\rm{e}}^4}} \right)\cdot\left( {3,4} \right) = 3 + 4{{\rm{e}}^4}$
