Answer
Not conservative.
Work Step by Step
A vector field will be conservative when $curl F=\nabla \times F=0$
$curl F= \nabla \times F=(\dfrac{\partial z}{\partial y}-\dfrac{\partial y}{\partial z})i+(\dfrac{\partial x}{\partial z}-\dfrac{\partial z}{\partial x})j+(\dfrac{\partial y}{\partial x}-\dfrac{\partial x}{\partial y})k$
This implies that
$curl F=(0-ye^z)i+(0-ze^x)j+(0-xe^y)k \ne 0$
Therefore, the given field is not conservative.