From Do Carmo’s Differential form and its application
To each tangent space $\mathbb R^3_p$ we can associate its dual space $(\mathbb R^3_p)^*$ which is the set of linear maps $\phi: \mathbb R^3_p \to \mathbb R$. A basis for $(\mathbb R^3_p)^*$ is obtained by taking $(dx_i)_p$ i=1,2,3 where $x_i:\mathbb R^3\to \mathbb R$ is the map which assigns to each point its $ i^{th}$ coordinate. The set {$(dx_i)_p$:$i=1,2,3$} is in fact the dual basis of {$(e_i)_p$} since
$(dx_i)_p(e_j)=\frac{\partial x_i}{\partial x_j}=0$ if $i=j$ and $ 1$ if $i\neq j$.
First thing I want to know what is this $(dx_i)_p$?
And Second how we got $(dx_i)_p(e_j)=\frac{\partial x_i}{\partial x_j}$
$(dx_i)_p$ is the Jacobian of the map $x_i$ at the point $p$. To see how to compute this, consider the map $$x_1 : \mathbb{R}^3 \to \mathbb{R}$$ $$(x,y,z) \mapsto x.$$ Then we have $$ (dx_1)_p = \begin{bmatrix} \frac{\partial}{\partial x}(x)|_p & \frac{\partial}{\partial y}(x)|_p & \frac{\partial}{\partial z}(x)|_p \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \end{bmatrix}. $$ Similarly, $$(dx_2)_p = \begin{bmatrix} 0 & 1 & 0 \end{bmatrix}$$ $$(dx_3)_p = \begin{bmatrix} 0 & 0 & 1 \end{bmatrix}.$$ Now, the maps $(dx_i)_p$ clearly define a linear map from $\mathbb{R}^3$ to $\mathbb{R}$ given by left multiplying any vector in $\mathbb{R}^3$ by the matrices above. These give a basis for the dual space $(\mathbb{R}^3_p)^*$ since any linear map $\phi: \mathbb{R}^3 \to \mathbb{R}$ takes the form $$\phi(x,y,z) = \alpha x + \beta y + \gamma z$$ for some $\alpha, \beta, \gamma \in \mathbb{R}$. Hence, we can write $$\phi = \alpha (dx_1)_p + \beta (dx_2)_p + \gamma (dx_3)_p.$$
The equation $(dx_i)_p(e_j) = \frac{\partial x_i}{\partial x_j}$ follows directly from our expressions for $(dx_i)_p$ above.