I am trying to prove this:
$\big[(\vec{A}\cdot\nabla)\vec{B}\big]_j=\sum_{k=1}^3 \Big[ \frac{A_k}{h_k}\frac{\partial B_j}{\partial q_k}+\frac{B_k}{h_k h_j}(A_j\frac{\partial h_j}{\partial q_k}-A_k\frac{\partial h_k}{\partial q_j}) \Big]$
as found on MathWorld. Here $h_i$ are Lamé coefficients (the square root of the diagonal elements of the metric tensor) and $q_i$ are the curvilinear coordinates. I denote the basis fields in those coordinates as $\vec{e}_i$ and $A_i$ and $B_i$ the respective components of the vectors above.
I tried to do it directly by writing:
$ \big[(\vec{A}\cdot\nabla)\vec{B}\big] = \sum_{j=1}^3\sum_{k=1}^3 \big((A_k \vec{e}_k) \cdot \nabla_q)\big) B_j\vec{e}_j $
where I assume: $ \nabla_q = \sum_{i=1}^3\frac{1}{h_i}\vec{e}_i\frac{\partial}{\partial q_i} $
which leads to:
$ \begin{align} \big[(\vec{A}\cdot\nabla)\vec{B}\big] &= \sum_{j=1}^3. \sum_{k=1}^3 \big((A_k \vec{e}_k) \cdot \nabla_q)\big) B_j\vec{e}_j \\ &= \sum_{j=1}^3\sum_{k=1}^3\Big[\frac{A_k}{h_k}\frac{\partial B_j}{\partial q_k}\vec{e}_j + \frac{A_k B_j}{h_k}\frac{\partial \vec{e}_j}{\partial q_k}\Big] \end{align} $
The first term is correct but I don't know what to do with the second one. I tried writing it as $\partial_k\vec{e_j} = \partial_k (\frac{1}{h_j}\frac{\partial \vec{x}}{\partial q_j})$ but i still end up with additional $\partial_j \vec{e_k}$, furthermore, in the resulting formula the second component in the sum has $B_k$ component and mixed $A_j$ and $A_k$ components so I am starting to doubt that there is a problem in the approach itself or maybe the nabla is written incorrectly?