#### Answer

Not conservative

#### Work Step by Step

A vector field is said to be conservative when $ curl F=\nabla \times F=0$
Now, $$ curl F= \nabla \times F \\=(\dfrac{\partial z}{\partial y}-\dfrac{\partial y}{\partial z}) \space i+(\dfrac{\partial x}{\partial z}-\dfrac{\partial z}{\partial x}) \space j+(\dfrac{\partial y}{\partial x}-\dfrac{\partial x}{\partial y}) \space k $$
From the given equation, we have
$ curl F=(0-ye^z)i+(0-ze^x)j+(0-xe^y)k \ne 0$
Therefore, the given field is not conservative.