Answer
The vector field is not conservative and thus, the line integral is not path independent.
Work Step by Step
Here, we have $P=f_x=\dfrac{\partial f}{\partial x};Q=f_y=\dfrac{\partial f}{\partial y}$ and $R=f_z=\dfrac{\partial f}{\partial z}$
Now, $\dfrac{\partial P}{\partial y}=\dfrac{\partial}{\partial y}(\dfrac{\partial f}{\partial x})=\dfrac{\partial^2 f}{\partial x \partial y}=\dfrac{\partial Q}{\partial x}$
Thus, we have $\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$
Now, $\dfrac{\partial P}{\partial z}=\dfrac{\partial}{\partial z}(\dfrac{\partial f}{\partial y})=\dfrac{\partial^2 f}{\partial x \partial z}=\dfrac{\partial R}{\partial x}$
Thus, we have $\dfrac{\partial P}{\partial z}=\dfrac{\partial R}{\partial x}$
Now, $\dfrac{\partial Q}{\partial z}=\dfrac{\partial}{\partial z}(\dfrac{\partial f}{\partial y})=\dfrac{\partial^2 f}{\partial z \partial y}=\dfrac{\partial R}{\partial y}$
Thus, we have $\dfrac{\partial Q}{\partial z}=\dfrac{\partial R}{\partial y}$
In the given problem, we have $\dfrac{\partial P}{\partial z}=0; \dfrac{\partial R}{\partial x}=yz$
Therefore, $\dfrac{\partial P}{\partial z} \neq \dfrac{\partial R}{\partial x}$
Hence, the given vector field is not conservative and thus, the line integral is not path independent.
