#### Answer

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
$$ curl F=-yz \times (x+yz)^{-2}+(x+yz)^{-1}+yz \times (x+yz)^{-2}-(x+yz)^{-1}-y(x+yz)^{-2}+y (x+yz)^{-2}+z \times (x+yz)^{-2}-z (x+yz)^{-2} \\= 0$$
Thus, the given field $ F $ is conservative.