Let us consider the transformation of rectangular coordinates $(x,y,z)$ into an orthogonal coordinates system $(u_1,u_2,u_3)$
Let $r = xi + yj + zk$ be the general position vector that can be written as $r=r(u_1,u_2,u_3)$; Now unit tangent vector in the direction of $u_1$ can be written as $$e_1=\frac{\frac{\partial r}{\partial u_1}}{\lvert \frac{\partial r}{\partial u_1} \rvert}\\Or,\ \ \ e_1=\frac{\frac{\partial r}{\partial u_1}}{h_1} \ \ \ \ where \ \ \ h_1=\frac{\partial r}{\partial u_1}$$
Similarly $e_2$ and $e_3$ can be written.
Now I'm trying to follow the proof of showing that $\frac{\partial r}{\partial u_1}, \frac{\partial r}{\partial u_2},\frac{\partial r}{\partial u_3}$ and $\nabla u_1,\nabla u_2,\nabla u_3,$ form a reciprocal system of vectors. But the first step in the proof is: $$\nabla u_1 = \frac{e_1}{h_1}$$
How is this expression obtained?
My attempt:
$$\nabla u_1 = \frac{\partial u_1}{\partial x}i + \frac{\partial u_1}{\partial y}j + \frac{\partial u_1}{\partial z}k$$
This doesn't lead me to involve $h_1$ term. So alternately,
$$\nabla u_1 = \frac{\partial u_1}{\partial u_1} e_1 + \frac{\partial u_1}{\partial u_2}e_2 + \frac{\partial u_1}{\partial u_3}e_3 = e_1 + 0 + 0 = e_1$$
I'm missing something conceptual. Kindly clarify.