We know that curl $\vec{F}$ is $$\vec{\nabla} \times \vec{F} = \langle \frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\rangle.$$ Is it true that $$ \textrm{div}(\textrm{curl} \, \vec{F}) = \vec{\nabla} \cdot (\vec{\nabla} \times \vec{F})= 0?$$ Assume that the scalar functions $P$, $Q$, and $R$ are so smooth that their mixed second partial derivatives are equal, like $$\frac{\partial^2 P}{\partial x \partial y}=\frac{\partial^2 P}{\partial y \partial x}.$$