Answer
${\bf{F}}$ is not conservative.
Work Step by Step
Assuming that ${\bf{F}}$ is a vector field on a simply connected domain ${\cal D}$, we check if ${\bf{F}} = \left( {{F_1},{F_2},{F_3}} \right) = \left( {\cos \left( {xz} \right),\sin \left( {yz} \right),xy\sin z} \right)$ satisfies the cross-partials condition:
$\dfrac{{\partial {F_1}}}{{\partial y}} = \dfrac{{\partial {F_2}}}{{\partial x}}$, ${\ \ \ }$ $\dfrac{{\partial {F_2}}}{{\partial z}} = \dfrac{{\partial {F_3}}}{{\partial y}}$, ${\ \ \ }$ $\dfrac{{\partial {F_3}}}{{\partial x}} = \dfrac{{\partial {F_1}}}{{\partial z}}$
We get
$0=0$, ${\ \ \ }$ $y\cos \left( {yz} \right) = x\sin z$, ${\ \ \ }$ $y\sin z = - x\sin \left( {xz} \right)$
There are inconsistencies, so we conclude that ${\bf{F}} = \left( {{F_1},{F_2},{F_3}} \right) = \left( {\cos \left( {xz} \right),\sin \left( {yz} \right),xy\sin z} \right)$ does not satisfy the cross-partials condition. Therefore, ${\bf{F}}$ is not conservative.
