I was wondering if the following step is correct. Suppose I have two vectors $\vec u$ and $\vec v$
$$\vec u \times (\nabla \times \vec v)=-(\nabla \times \vec v)\times\vec u \stackrel{?}{=} -(-(\vec v\times \nabla )\times \vec u)=(\vec v\times \nabla)\times \vec u $$ I think the gradient operator is not commutative under the cross product, as it is not commutative under the dot product, but then, how can I switch the $\nabla$ and $\vec v$?
The identity cannot be true, because the LHS contains partial derivatives of the fuction $v$, whereas the RHS contains partial derivatives of the function $u$. This also puts a big question mark on any operation that would be supposed to "switch" $\nabla$ and $v$.