Answer
The vector field is not conservative.
Work Step by Step
We have ${\bf{F}} = \left( {{F_1},{F_2},{F_3}} \right) = \left( {xyz,\frac{1}{2}{x^2}z,2{z^2}y} \right)$.
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
$xz = xz$, ${\ \ \ \ }$ $\frac{1}{2}{x^2} = 2{z^2}$, ${\ \ \ \ }$ $0 = xy$
Since ${\bf{F}}$ does not satisfy the cross-partials condition, therefore by Theorem 4 in Section 17.3, ${\bf{F}}$ is not conservative.
