Using Feynman notation $\nabla \times (\mathbf{A} \times \nabla) =\nabla_A \times (\mathbf{A} \times \nabla) +\nabla_\nabla \times (\mathbf{A} \times \nabla)$ but I have a problem while expanding the term $\nabla_A \times (\mathbf{A} \times \nabla) $ because the last nabla isn't acting on anything so I can't get the first term of the BAC rule.
I guess I solved it on my own, but I want to know if there is any mistake:
$$\nabla_A \times (\mathbf{A} \times \nabla)=(\nabla_{none} \cdot \nabla)\mathbf{A}-(\nabla \cdot \mathbf{A}) \nabla $$
In the special case $\mathbf{r}=\mathbf{A}$
\begin{align*} \nabla_r \times (\mathbf{r} \times \nabla)&=(\nabla_{none} \cdot \nabla)\mathbf{r}-(\nabla \cdot \mathbf{r}) \nabla \\ &=(\partial{x}\partial{x}+\partial{y}\partial{y}+\partial{z}\partial{z})(x,y,z) - 3\nabla\\ &=(\partial{y},\partial{y},\partial{z}) -3\nabla\\ &=-2\nabla\\ \end{align*}
Here I used just one of pair the partial derivatives since one doesn't act on anything.
This kind of problems always get much easier if you use Levi-Civita symbol, i.e., you can rewrite the equation as
$\epsilon_{ijk}\partial_{j}(\epsilon_{klm}A_{l}\partial_{m})$.
Then using the identity $\epsilon_{ijk}\epsilon_{klm}= \delta_{il}\delta_{jm}-\delta_{im}\delta_{jl},$ you obtain
$\partial_{j}(A_{i}\partial_{j}) -\partial_{l}(A_{l}\partial_{i})$.