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