## Thomas' Calculus 13th Edition

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.