Orthogonal curvilinear coordinates (derivatives of unit vectors) Suppose that $\{u_i\}_{1\le i\le 3}$ is a set of orthogonal curvilinear coordinates with unit vectors $\{\mathbf{\hat{e}_i}\}_{1\le i\le 3}$. I proved that
$$\frac{\partial \mathbf{\hat{e}_i}}{\partial u_j} = \frac{\mathbf{\hat{e}_j}}{h_i}\frac{\partial h_j}{\partial u_i},\tag{1}$$
where $h_i$ is a scale factor such that for a position vector $\mathbf r$ we have $\dfrac{\partial \mathbf r}{\partial u_i}= h_i \mathbf{\hat{e}_i}$. Eq. $(1)$ is valid for $i\neq j$. I'm trying to prove that
$$\frac{\partial \mathbf{\hat{e}_i}}{\partial u_i} = -\sum\limits_{k\neq i}\frac{\mathbf{\hat{e}_k}}{h_k}\frac{\partial h_i}{\partial u_k}.$$
First of all I find it strange that the variation of $\mathbf{\hat{e}_i}$ with respect to the $i$-th coordinate is not pointing towards $\mathbf{\hat{e}_i}$. For instance in Eq. (1) we have that $\dfrac{\partial \mathbf{\hat{e}_i}}{\partial u_j}\parallel \mathbf{\hat{e}_j}$, which makes sense. And secondly this is what I tried to prove the equality (I followed some steps that were suggested, I will mark these steps with $\overset{*}{=}$):
\begin{align}
\mathbf{\hat{e}_i} = \mathbf{\hat{e}_j}\times \mathbf{\hat{e}_k} = \sum\limits_{i=1}^3\epsilon_{ijk}\mathbf{\hat{e}_i} \overset{*}{=}\frac12\sum\limits_{j,k=1}^3\epsilon_{ijk}\mathbf{\hat{e}_j}\times \mathbf{\hat{e}_k},
\end{align}
where $\epsilon_{ijk}$ is the Levi-Civita tensor.
\begin{align}
\frac{\partial \mathbf{\hat{e}_i}}{\partial u_i} &= \frac12\sum\limits_{j,k=1}^3\epsilon_{ijk}\frac{\partial}{\partial u_i}(\mathbf{\hat{e}_j}\times \mathbf{\hat{e}_k}) = \frac12\sum\limits_{j,k=1}^3\epsilon_{ijk}\left(\frac{\partial\mathbf{\hat{e}_j}}{\partial u_i}\times \mathbf{\hat{e}_k}  + \frac{\partial\mathbf{\hat{e}_k}}{\partial u_i}\times \mathbf{\hat{e}_j}\right)\\
&\overset{*}{=} \frac12\sum\limits_{j,k=1}^3\epsilon_{ijk}\left(\frac{\mathbf{\hat{e}_i}}{h_j}\frac{\partial h_i}{\partial u_j}\times \mathbf{\hat{e}_k}  + \frac{\mathbf{\hat{e}_i}}{h_k}\frac{\partial h_i}{\partial u_k}\times \mathbf{\hat{e}_j}\right)=\frac12\sum\limits_{j,k=1}^3\epsilon_{ijk}\left(-\frac{\mathbf{\hat{e}_j}}{h_j}\frac{\partial h_i}{\partial u_j}  + \frac{\mathbf{\hat{e}_k}}{h_k}\frac{\partial h_i}{\partial u_k}\right).
\end{align}
I feel so close...
I have the hint that one should get to this result:
$$\frac{\partial \mathbf{\hat{e}_i}}{\partial u_i} = \sum_{jkl}\epsilon_{ijk}\epsilon_{ilj}\frac{\mathbf{\hat{e}_l}}{h_k}\frac{\partial h_i}{\partial u_k}.$$
But I don't know how to get to that. Any help is very much appreciated. Thanks.
 A: Not familar with Levi-Civita, but here is an alternative proof:
\begin{equation}
\mathbf{\hat e_i} \cdot \mathbf{\hat e_i}=1 \\
\frac{\partial\mathbf{\hat e_i}}{\partial u_i}\cdot \mathbf{\hat e_i} = 0
\end{equation}
Thus
$\frac{\partial \mathbf{\hat e_i}}{\partial u_i}$ has no $i$ component.
For $i\neq j$
\begin{equation}
\mathbf{\hat e_i} \cdot \mathbf{\hat e_j}=0 \\
\frac{\partial \mathbf{\hat e_i}}{\partial u_i}\cdot \mathbf{\hat e_j} = -\frac{\partial \mathbf{\hat e_j}}{\partial u_i}\cdot \mathbf{\hat e_i}
\end{equation}
So the $j$ component of $\frac{\partial\mathbf{\hat e_i}}{\partial u_i}$ is $- \mathbf{\hat e_i}\cdot \frac{\partial\mathbf{\hat e_j}}{\partial u_i}$.
\begin{equation}
- \mathbf{\hat e_i}\cdot \frac{\partial\mathbf{\hat e_j}}{\partial u_i}\\
=-\frac{\partial \mathbf{r}}{h_i\partial u_i}\cdot\frac{\partial}{\partial u_i}(\frac{\partial \mathbf{r}}{h_j\partial u_j})\\
=-\frac{\partial \mathbf{r}}{h_i\partial u_i}\cdot(\frac{\partial}{\partial u_i}(\frac{1}{h_j})\frac{\partial \mathbf{r}}{\partial u_j} + \frac{1}{h_j}\frac{\partial^2 \mathbf{r}}{\partial u_i \partial u_j})
\end{equation}
The first item is in the direction of $j$, thus
\begin{equation}
- \mathbf{\hat e_i}\cdot \frac{\partial\mathbf{\hat e_j}}{\partial u_i}\\
=-\frac{1}{h_i h_j}\frac{\partial \mathbf{r}}{\partial u_i}\cdot\frac{\partial^2 \mathbf{r}}{\partial u_i \partial u_j}\\
=-\frac{1}{2h_i h_j}(\frac{\partial}{\partial u_j}(\frac{\partial \mathbf{r}}{\partial u_i}\cdot \frac{\partial \mathbf{r}}{\partial u_i}) )\\
=-\frac{1}{2h_i h_j}\frac{\partial}{\partial u_j}h_i^2\\
=-\frac{1}{ h_j}\frac{\partial}{\partial u_j}h_i
\end{equation}
Thus,
\begin{equation}
\frac{\partial \mathbf{\hat e_i}}{\partial u_i}\cdot \mathbf{\hat e_j}=-\frac{1}{ h_j}\frac{\partial}{\partial u_j}h_i
\end{equation}
Similarly,
\begin{equation}
\frac{\partial \mathbf{\hat e_i}}{\partial u_i}\cdot \mathbf{\hat e_k}=-\frac{1}{ h_k}\frac{\partial}{\partial u_k}h_i
\end{equation}
