understanding Do Carmo's Differential form From Do Carmo’s Differential form and its application
To each tangent space $R^3_p$ we can associate its dual space $(R^3_p)^*$ which is the set of linear maps $\phi: R^3_p -> R$.A basis for $(R^3_p)^*$ is obtained bytaking $(dx_i)_p$ i=1,2,3 where $x_i:R^3->R$ 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 {$(ei)_p$} since
$(dx_i)_p(e_j)=\frac{\partial x_i}{\partial x_j}$= 0 if i=j $and $ 1 if $ i\neq j$.
Ist thing I want to know what is this $(dxi)_p$?
And Second how we got $(dxi)_p(e_j)=\frac{\partial x_i}{\partial x_j}$
 A: $(dx_i)_p$ 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.
