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