Does such a vector field exist?

Question

Does there exist a velocity field $\mathbf{u}(x,t)\in \mathbb{R}^3$ such that $$\text{Div}\begin{bmatrix} \mathbf{u}\cdot \nabla w_1\\ \mathbf{u}\cdot \nabla w_2\\ \mathbf{u}\cdot \nabla w_3\\ \end{bmatrix} = \text{Div}\begin{bmatrix} \mathbf{w}\cdot \nabla u_1\\ \mathbf{w}\cdot \nabla u_2\\ \mathbf{w}\cdot \nabla u_3 \end{bmatrix} $$ where $$\mathbf{w} = \nabla \times u = \operatorname{Curl}(\text{u})$$ and in this context $\mathbf{w} = (w_1,w_2,w_3)$. Let $D_\textbf{v}(f)$ denote the directional derivative of $f$ in the direction of the vector field $\mathbf{v}$ then for each point in time $t>0$ we have $$\text{Div}\begin{bmatrix} D_\mathbf{u}\left(w_1\right)\\ D_\mathbf{u}\left(w_2\right)\\ D_\mathbf{u}\left(w_3\right) \end{bmatrix} = \text{Div}\begin{bmatrix} D_\mathbf{w}\left(u_1\right)\\ D_\mathbf{w}\left(u_2\right)\\ D_\mathbf{w}\left(u_3\right) \end{bmatrix}. $$

Using Einstein notation we can write this simply as $$\dfrac{\partial}{\partial x_i} D_\mathbf{u}\left(w_i\right) = \dfrac{\partial}{\partial x_i}D_\mathbf{w}\left(u_i\right)$$ where in this context of course $\dfrac{\partial }{\partial x_1} = \dfrac{\partial }{\partial x}$, $\dfrac{\partial }{\partial x_2} = \dfrac{\partial }{\partial y}$, and $\dfrac{\partial }{\partial x_3} = \dfrac{\partial }{\partial z}$. Intuitively, this suggest that $\mathbf{w}$ and $\mathbf{u}$ are the same up to some constant vector $\xi$. Can we show that $\mathbf{u} = \mathbf{w} + \xi$? This not general true for arbitrary fields unless their curl is the same.