How is it $\nabla.(\nabla u_2 \times \nabla u_3)=0$? 
I think
Since $\nabla u_2 \times \nabla u_3=\dfrac{\vec{e_1}}{h_2h_3} ,$ then $ \nabla\cdot(\nabla u_2 \times \nabla u_3)=\dfrac{1}{h_1} \dfrac{\partial}{\partial u_1}\bigg(\dfrac{1}{h_1h_2} \bigg)$
but
How is it $\nabla\cdot(\nabla u_2 \times \nabla u_3)=0$ ?
 A: Using the definition
$$
\nabla \cdot (A_1\mathbf{e}_1) = \frac{1}{h_1h_2h_3}\frac{\partial}{\partial u_1}(A_1h_2h_3)
$$
we can see that $A_1 = \frac{1}{h_2h_3}$
$$
\nabla \cdot (\frac{1}{h_2h_3} \mathbf{e}_1)  = \frac{1}{h_1h_2h_3}\frac{\partial}{\partial u_1}(\frac{1}{h_2h_3} \times h_2h_3)=\frac{1}{h_1h_2h_3}\frac{\partial}{\partial u_1} 1 = 0
$$
A: This is a standard vector calculus identity:
$$
\nabla\cdot(\nabla f \times \nabla g) = (\nabla\times\nabla f)g - (\nabla\times\nabla g)f = 0.
$$
A: Here's a graphical derivation of the identity $\nabla\cdot(\nabla f\times \nabla g)=0$, following the notation in the fun paper Boosting Vector Calculus with the Graphical Notation.
In the following, the left-hand side represents $\nabla\cdot(\nabla f\times \nabla g)$, and to get the right-hand side we apply the product rule for the outer bubble.

We need the following identity:

This is a bubble swap (a change in the order of the variables of differentiation) followed by swapping the inputs of the cross product, which negates the expression.  Since this is saying the left-hand side diagram is negative of itself, it's equal to zero.
Going back to the original diagrams, we can see in the sum that the first term has this diagram with $h=f$ and the second has it with $h=g$, so both terms are zero. Therefore $\nabla\cdot(\nabla f\times \nabla g)=0$.

By the way, the first diagram with $h$ represents the following vector:
$$\sum_{i,j} \frac{\partial^2 h}{\partial u_i\, \partial u_j}(\mathbf{e}_i\times \mathbf{e}_j)$$
where $h$ is a function of $(u_1,u_2,u_3)$.  The graphical proof is that
$$\sum_{i,j} \frac{\partial^2 h}{\partial u_i\, \partial u_j}(\mathbf{e}_i\times \mathbf{e}_j) = - \sum_{i,j} \frac{\partial^2 h}{\partial u_j\, \partial u_i}(\mathbf{e}_j\times \mathbf{e}_i).$$
A: Using Levi—Civita symbol $\varepsilon^{ijk}$ (and assuming $\partial_i(\partial_jf) =\partial_j(\partial_if)$ and $\partial_i(\partial_jg) =\partial_j(\partial_ig)$) one can get
$$
\nabla\cdot(\nabla f \times \nabla g) = \varepsilon^{ijk}\partial_i (\partial_jf \partial_k g) = \underbrace{\varepsilon^{ijk}\partial_i(\partial_jf)\partial_kg}_0 + \underbrace{\varepsilon^{ijk}\partial_jf\partial_i(\partial_kg)}_0 = 0.
$$

PS. I used here the following trick (and the property $\varepsilon^{ijk} = -\varepsilon^{jik}$)
$$
\varepsilon^{ijk}\partial_i(\partial_jf)\partial_k g = 
\tfrac12 \varepsilon^{ijk}\partial_i(\partial_jf)\partial_k g
+\tfrac12 \varepsilon^{ijk}\partial_i(\partial_jf)\partial_k g = 
\tfrac12 \varepsilon^{ijk}\partial_i(\partial_jf)\partial_k g
+\tfrac12 \varepsilon^{jik}\partial_j(\partial_if)\partial_k g = 
\tfrac12 \varepsilon^{ijk}\partial_i(\partial_jf)\partial_k g
-\tfrac12 \varepsilon^{ijk}\partial_i(\partial_jf)\partial_k g = 0.
$$
Same with the term $\varepsilon^{ijk}\partial_jf\partial_i(\partial_kg)$.
