Answer
The vector field is not conservative.
Work Step by Step
We have ${\bf{F}} = \left( {{F_1},{F_2},{F_3}} \right) = \left( {x{{\rm{e}}^{2x}},y{{\rm{e}}^{2z}},z{{\rm{e}}^{2y}}} \right)$.
Let us 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
$0=0$, ${\ \ \ \ }$ $2y{{\rm{e}}^{2z}} = 2z{{\rm{e}}^{2y}}$, ${\ \ \ \ }$ $0=0$
Since ${\bf{F}}$ does not satisfy the cross-partials condition, therefore by Theorem 4 in Section 17.3, ${\bf{F}}$ is not conservative.
