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

Using Eisenstein notation we can write this simply as $$\dfrac{\partial}{\partial x_i} D_\textbf{u}\left(w_i\right) =  \dfrac{\partial}{\partial x_i}D_\textbf{w}\left(u_i\right)$$ 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 $\textbf{w}$ and $\textbf{u}$ are the same up to some constant vector $\xi$. Can we show that $\textbf{u} = \textbf{w} + \xi$? This not general true for arbitrary fields unless their curl is the same.