Answer
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 2\left( {{{\rm{e}}^2} - {{\rm{e}}^{ - 2}}} \right) - \left( {{\rm{e}} - {{\rm{e}}^{ - 1}}} \right)$
Work Step by Step
We have the vector field ${\bf{F}}\left( {x,y} \right) = \left( {3z{y^{ - 1}},4x, - y} \right)$ and the parametrization ${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^t},{{\rm{e}}^t},t} \right)$ for $ - 1 \le t \le 1$.
So, we obtain
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {3t{{\rm{e}}^{ - t}},4{{\rm{e}}^t}, - {{\rm{e}}^t}} \right)$
$d{\bf{r}} = {\bf{r}}'\left( t \right)dt$
$d{\bf{r}} = \left( {{{\rm{e}}^t},{{\rm{e}}^t},1} \right)dt$
Evaluate the dot product ${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right)dt$:
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right)dt = \left( {3t{{\rm{e}}^{ - t}},4{{\rm{e}}^t}, - {{\rm{e}}^t}} \right)\cdot\left( {{{\rm{e}}^t},{{\rm{e}}^t},1} \right)dt$
$ = \left( {3t + 4{{\rm{e}}^{2t}} - {{\rm{e}}^t}} \right)dt$
Evaluate $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$ using Eq. (8):
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_a^b {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_{ - 1}^1 \left( {3t + 4{{\rm{e}}^{2t}} - {{\rm{e}}^t}} \right){\rm{d}}t$
$ = \left( {\left( {\frac{3}{2}{t^2} + 2{{\rm{e}}^{2t}} - {{\rm{e}}^t}} \right)|_{ - 1}^1} \right) = \frac{3}{2} + 2{{\rm{e}}^2} - {\rm{e}} - \frac{3}{2} - 2{{\rm{e}}^{ - 2}} + {{\rm{e}}^{ - 1}}$
$ = 2{{\rm{e}}^2} - {\rm{e}} - 2{{\rm{e}}^{ - 2}} + {{\rm{e}}^{ - 1}} = 2\left( {{{\rm{e}}^2} - {{\rm{e}}^{ - 2}}} \right) - \left( {{\rm{e}} - {{\rm{e}}^{ - 1}}} \right)$
So, $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 2\left( {{{\rm{e}}^2} - {{\rm{e}}^{ - 2}}} \right) - \left( {{\rm{e}} - {{\rm{e}}^{ - 1}}} \right)$.
