Answer
(a) We show that ${\bf{F}}$ is not conservative.
(b) Without explicitly computing the integral, we show that
$\mathop \smallint \limits_{{C_2}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 0$
Work Step by Step
(a) Write ${\bf{F}} = \left( {{F_1},{F_2},{F_3}} \right) = \left( {9y - {y^3},{{\rm{e}}^{\sqrt y }}\left( {{x^2} - 3x} \right),0} \right)$.
We check if ${\bf{F}}$ satisfies the cross-partials condition:
$\frac{{\partial {F_1}}}{{\partial y}} = \frac{{\partial {F_2}}}{{\partial x}}$, ${\ \ \ \ }$ $\frac{{\partial {F_2}}}{{\partial z}} = \frac{{\partial {F_3}}}{{\partial y}}$, ${\ \ \ \ }$ $\frac{{\partial {F_3}}}{{\partial x}} = \frac{{\partial {F_1}}}{{\partial z}}$
We get
$9 - 3{y^2} = {{\rm{e}}^{\sqrt y }}\left( {2x - 3} \right)$, ${\ \ \ \ }$ $0=0$, ${\ \ \ \ }$ $0=0$
Since ${\bf{F}}$ does not satisfy the cross-partials condition, therefore by Theorem 4 in Section 17.3, ${\bf{F}}$ is not conservative.
(b) Referring to Figure 1(B), we can consider the curve ${C_2}$ consists of 4 segments: along the $x$-axis, along the line segment $x=3$, along the line segment $y=3$, and along the $y$-axis. Thus, the line integral along ${C_2}$ is the sum of the line integrals of these segments:
$\mathop \smallint \limits_{{C_1}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_{x - axis}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} + \mathop \smallint \limits_{x = 3}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} + \mathop \smallint \limits_{y = 3}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} + \mathop \smallint \limits_{y - axis}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$
We have ${\bf{F}}\left( {x,y} \right) = \left( {9y - {y^3},{{\rm{e}}^{\sqrt y }}\left( {{x^2} - 3x} \right)} \right)$.
1. along the $x$-axis
In this case, $y=0$, so the vector field becomes
${\bf{F}}\left( {x,0} \right) = \left( {0,{x^2} - 3x} \right)$
Notice that ${\bf{F}}\left( {x,0} \right)$ is parallel to the $y$-axis, thus it is perpendicular to the $x$-axis. Therefore, ${\bf{F}}\left( {x,0} \right)\cdot d{\bf{r}} = 0$. So,
$\mathop \smallint \limits_{x - axis}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 0$
2. along the line segment $x=3$
In this case, $x=3$, so the vector field becomes
${\bf{F}}\left( {3,y} \right) = \left( {9y - {y^3},0} \right)$
Notice that the vector ${\bf{F}}\left( {3,y} \right)$ points in the direction that is parallel to the $x$-axis, thus it is perpendicular to the line segment. Therefore, ${\bf{F}}\left( {3,y} \right)\cdot d{\bf{r}} = 0$. So,
$\mathop \smallint \limits_{x = 3}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 0$
3. along the line segment $y=3$
In this case, the vector field becomes
${\bf{F}}\left( {x,3} \right) = \left( {0,{{\rm{e}}^{\sqrt 3 }}\left( {{x^2} - 3x} \right)} \right)$
Since the the vector ${\bf{F}}\left( {x,3} \right)$ points in the direction that is parallel to the $y$-axis, thus it is perpendicular to the line segment. Therefore, ${\bf{F}}\left( {x,3} \right)\cdot d{\bf{r}} = 0$. So,
$\mathop \smallint \limits_{y = 3}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 0$
4. along the $y$-axis
In this case, $x=0$, so the vector field becomes
${\bf{F}}\left( {0,y} \right) = \left( {9y - {y^3},0} \right)$
We have here the vector ${\bf{F}}\left( {0,y} \right)$ that is parallel to the $x$-axis, thus it is perpendicular to the $y$-axis. Therefore, ${\bf{F}}\left( {0,y} \right)\cdot d{\bf{r}} = 0$. So,
$\mathop \smallint \limits_{y - axis}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 0$
Hence
$\mathop \smallint \limits_{{C_2}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_{x - axis}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} + \mathop \smallint \limits_{x = 3}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} + \mathop \smallint \limits_{y = 3}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} + \mathop \smallint \limits_{y - axis}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 0$
