I am reading a "Primer on Tensor Calculus" by Clark from here.
In Def. 3.5 he defines a covariant basis vector as: Let $\vec{r}_x$ be a displacement vector whose components are expressed in terms of the coordinate system $x^i$. Then the covariant basis vector $\vec{e}_i$ is defined to be:
$$ \begin{align} \vec{e}_i \equiv \frac{\mathrm{d}\vec{r}_x}{\mathrm{d}x^i} \end{align} $$
Then he invokes that because the differential displacement vector is: $$ \begin{align} \mathrm{d}\vec{r} = \sum_{i} h_{(i)}\mathrm{d}x^i\hat{e}_{(i)} \end{align} $$ where $\hat{e}_{(i)}$ are unit "physical" vectors, it follows that:
$$ \begin{align} \mathbb{e}_i = h_{(i)}\hat{e}_{(i)} \end{align} $$
I don't understand what he means by the symbol $\frac{\mathrm{d}\vec{r}_x}{\mathrm{d}x^i}$. From what i remember, this should be a tangent to the $x^i$ coordinate line, but if that is so, I would write it as $\frac{\partial\vec{r}_x}{\partial x^i}$. The way he writes it it looks like some division but then I am not sure how to show the covariant character of this vector.
Secondly, how does he get $\mathbb{e}_i = h_{(i)}\hat{e}_{(i)}$ just using the definition and the differential displacement?