Answer
The line integral is not path independent.
Work Step by Step
Here, we have $\dfrac{\partial P}{\partial z}=\dfrac{\partial}{\partial z}(\dfrac{\partial f}{\partial y})$
$\implies \dfrac{\partial^2 f}{\partial x \partial z}=\dfrac{\partial R}{\partial x}$
This shows that $\dfrac{\partial P}{\partial z}=\dfrac{\partial R}{\partial x}$
and $\dfrac{\partial Q}{\partial z}=\dfrac{\partial}{\partial z}(\dfrac{\partial f}{\partial y})$
$\implies\dfrac{\partial^2 f}{\partial z \partial y}=\dfrac{\partial R}{\partial y}$
Thus, we have $\dfrac{\partial Q}{\partial z}=\dfrac{\partial R}{\partial y}$
Since, $\dfrac{\partial P}{\partial z}=0$ and $ \dfrac{\partial R}{\partial x}=yz$
This implies that $\dfrac{\partial P}{\partial z} \neq \dfrac{\partial R}{\partial x}$
Hence, we have the vector field is not conservative and so, the line integral is not path independent.
