I have to prove that,
$$\nabla \cdot (A_1\mathbf e_1) = \frac{1}{h_1h_2h_3}\frac{\partial (A_1h_2h_3)}{\partial u_1} $$
My approach:
$$\begin{align} \nabla \cdot (A_1\mathbf e_1) &= \nabla \cdot (A_1 h_1 \nabla u_1)\\ &= \underbrace{\nabla (A_1h_1) \cdot \nabla u_1} + \underbrace{A_1h_1 \nabla \cdot \nabla u_1}\\ &= \underbrace {\frac{1}{h_1^{2}} \frac{\partial(A_1h_1)}{\partial u_1}} + \underbrace{\nabla^{2}u_1}\\ &= \underbrace {\frac{1}{h_1^{2}} \frac{\partial(A_1h_1)}{\partial u_1}} + \underbrace{\frac{1}{h_1h_2h_3}\frac{\partial(\frac{h_2h_3}{h_1})}{\partial u_1}}\\ \end{align} $$
Notes:
To expand the first step I used $\nabla \cdot (\phi \mathbf A) = \nabla \phi \cdot \mathbf A + \phi \nabla \cdot \mathbf A$
To expand the third step I used the expansion formula for Laplace operator i.e.,
\begin{eqnarray} \nabla ^2 \phi = \frac{1}{h_1 h_2 h_3}\left(\frac{\partial }{\partial u_1}\left[\frac{h_2h_3}{h_1}\frac{\partial \phi}{\partial u_1}\right] + \ldots \right) \end{eqnarray}
Is there any mistake in any of the above steps? Could it be simplified further from here?
Please note that I do have the right solution for which in the first step I'll have to substitute $\mathbf e_1=h_2h_3\nabla u_2 \times \nabla u_3$ and proceed.