Answer
a) $(x \cos xy-x \cos zx)i+(y \cos yz-y \cos xy)j+(z \cos zx-z \cos yz)k$
b) $0$
Work Step by Step
a) $curl F=(x \cos xy-x \cos zx)i+(y \cos yz-y \cos xy)j+(z \cos zx-z \cos yz)k$
or, $=(x \cos xy-x \cos zx)i+(y \cos yz-y \cos xy)j+(z \cos zx-z \cos yz)k$
b) $div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}$
This implies that
$div F=\dfrac{\partial (\sin yz)}{\partial x}+\dfrac{\partial (\sin zx)}{\partial y}+\dfrac{\partial (\sin xy)}{\partial z}$
or, $div F=0$
