Answer
As $\nabla \times \textbf F \ne 0$, so, $\textbf F$ is not conservative
Work Step by Step
$\textbf F(x,y) = \sin (yz) \textbf i +xz\cos (yz) \textbf j+xy \sin(yz)\textbf k $
$\nabla \times \textbf F =\begin{vmatrix}
\textbf i & \textbf j & \textbf k\\
\frac{\delta}{\delta x} & \frac{\delta}{\delta y}& \frac{\delta}{\delta z} \\
\sin (yz) & xz\cos (yz) & xy \sin(yz) \\
\end{vmatrix} \\
= [x\sin(yz)+xyz\cos(yz)-x\cos(yz)+xyz\sin(yz)]\textbf i + [y\cos(yz)-y\sin(yz)]\textbf j +[z\cos(yz) - y\cos(yz)] \textbf k\ne 0
\\$
As $\nabla \times \textbf F \ne 0$, so, $\textbf F$ is not conservative
