Suppose that $\phi$ is any trial function that is continuously differentiable in the closure of $D$ and vanishes on $\partial D$, then by the divergence theorem
\begin{align*}\int_D \left [ \frac{\partial w}{\partial x_j}\frac{\partial \phi}{\partial x_k} - \frac{\partial w}{\partial x_k}\frac{\partial \phi}{\partial x_j} \right]dx &= \int_D \phi \left [ \frac{\partial^2 w}{\partial x_j \partial x_k} - \frac{\partial ^2 w}{\partial x_k \partial x_j} \right]dx \\ &- \int_{\partial D} \phi \left [ \frac{\partial w}{\partial x_j} \frac{\partial x_k}{\partial v} - \frac{\partial w}{\partial x_k} \frac{\partial x_j}{\partial v} \right]dx \\ &= 0 \end{align*}
How can we get this equation by divergence theorem?
A simple corollary of the divergence theorem applied to the vector field $v=e_j \partial_k w \phi$ is $$ \int_D \nabla \cdot (e_j \partial_k w \phi) \;dx=\int_D \phi \partial_j \partial_k w + (\partial_k w)(\partial_j \phi)\;dx=\int_{\partial D} e_j \partial_k w \phi \cdot n \;ds=0 $$ where $n$ denotes the outer normal vector of $D$. The last equality is a consequence of $\phi$ vanishing on the boundary. Applying this formula to $v=e_k \partial_j w \phi$ as well, gives you $$ \int_D \phi \partial_j \partial_k w + (\partial_k w)(\partial_j \phi)\;dx+\int_D \phi \partial_k \partial_j w + (\partial_j w)(\partial_k \phi)\;dx=0 $$ Using Schwartzs theorem to commute partial derivatives and rearranging gives you: $$ \int_D (\partial_k w)(\partial_j \phi)-(\partial_j w)(\partial_k \phi)\;dx=\int_D (\partial_k\partial_j w-\partial_j \partial_k w)\;dx=\int_D (\partial_k\partial_j w-\partial_k \partial_j w)\;dx=0 $$