Orthornormal basis and Dual basis If $e_a$ is an orthonormal basis for vectors and $\theta^a$ the dual basis for coordinate vectors. How to prove that metric is expressed as $ds^2=\delta_{ab} \theta^a \theta^b$ and  
$e_{a}^{i}\theta_{j}^{a}=\delta_{j}^{i}$  and   $g^{ij}=\delta^{ab} e_{a}^{i} e_{b}^{i}$?
 A: I think there are some issues of notational confusion here which I want to make sure are sorted before answering. This is quite a long response; I apologize if some of it is off topic or rambly, but it's difficult to tell quite where you are confused exactly.
It looks like you're using abstract index notation for vectors and tensors and so forth (which is fine), but in that case the expression "$e_a$" denotes a single vector, not an orthonormal basis of vectors, which would look more like
$$
  e^1_a, e^2_a, \dots, e^n_a
$$
as suggested by the formulas below. (The "a" subscript isn't literally an index in the same way the "i" is an index in an expression like $e^i_a$. The "a" subscript just says "this is a $(0,1)$-tensor" -- a vector. Sometimes people will even write things like $(e_a)^i$ to try to be extra clear about this. I will attempt to be clear by using $a,b,c, \dots$ for abstract indicies -- which never get numbers fed into them -- and either $1,2, \dots,$ or $i,j,k,\dots$ for the normal kind of everyday indicies.) 
Likewise, a dual basis should be written
$$
  \theta^a_1, \theta^a_2, \dots, \theta^a_n.
$$
What does it mean for the $\theta^a_i$ to be dual to the $e^j_a$? By definition this means that
$$
  \theta^a_i e^j_a = \delta^j_i,
$$
which is your second equation. 
Your first equation is an expression for the metric tensor, which I'm not particularly happy writing as $ds^2$ in the abstract index notation because there aren't indicies indicating what kind of tensor it is. I'll write the metric tensor for now as $m^{ab}$; it's a $(2,0)$-tensor. (The notation $g^{ab}$ is more common, but you've written something funny using $g^{ij}$ in your third equation and I don't want to get mixed up with that.)
$m^{ab}$ is the metric tensor in the sense that given any two vectors $x_a$ and $y_a$, the inner product of $x_a$ and $y_a$ as determined by the metric $m^{ab}$ is
$$
    m^{ab} x_a y_b.
$$
What does it mean for the $e_a^i$ to be orthonormal? It means that when I take the inner product of $e_a^i$ and $e_a^j$ for any $i$ and $j$, I should get out $\delta^{ij}$. In equations,
$$
    m^{ab} e_a^i e_b^j = \delta^{ij}.
$$ 
Maybe this is what you're looking for in your third equation?
In any case, this equation uniquely determines $m^{ab}$ since the $e_a^i$ form a basis. But there's another easy-to-write-down $(2,0)$-tensor satisfying this equation, namely, the tensor
$$
   \delta^{kl} \theta^a_k \theta^b_l,
$$
which you should be able to check satisfies
$$
   \delta^{kl} \theta^a_k \theta^b_l e_a^i e_b^j = \delta^{ij}
$$
from which it follows (since the $e_a^i$ form a basis) that
$$
  m^{ab} = \delta^{kl} \theta^a_k \theta^b_l,
$$
which I think is what you're trying to claim with your first equation.
Finally, a warning, related to your third equation. Most mathematicians don't use abstract index notation (it's more of a physics thing), but you will see mathematicians using the symbol $g^{ij}$. When mathematicians use this symbol (in the context of Riemannian geometry), they almost always mean the following: there is, implicitly or explicitly, a coordinate chart on your manifold $M$ we are discussing with coordinates $x_1, \dots, x_n$ (these are functions on an open subset of $M$), and these give rise to a local frame
$$
   \frac{\partial}{\partial x_1} , \dots, \frac{\partial}{\partial x_n},
$$
which are now a collection of tangent vector fields on this open subset. (In abstract index notation you'd write these as
$$
   \Big( \frac{\partial}{\partial x_1} \Big)_a , \dots, \Big( \frac{\partial}{\partial x_n} \Big)_a,
$$
the "a" subscript just reminding us that these are vector fields.) In general, the vectors $\partial / \partial x_i$ will not be orthonormal; the symbol $g^{ij}$ denotes the inner product
$$
   g^{ij} = \Big\langle \frac{\partial}{\partial x_i}, \frac{\partial}{\partial x_j} \Big\rangle
$$
which in abstract index notation and using $m^{ab}$ for the metric tensor as above would be written
$$
   g^{ij} = m^{ab} \Big( \frac{\partial}{\partial x_i} \Big)_a \Big( \frac{\partial}{\partial x_j} \Big)_b.
$$
In either case $g^{ij}$ is, for any given $i,j$, a function on $M$ (as represented by the fact that it has no abstract indicies attached to it) which depends on the choice of coordinates $x_1, \dots, x_n$. These functions taken together carry all the data of the metric tensor, but in a non-coordinate-invariant way.
In particular, if you want to begin with a local orthonormal frame $e^i_a$ as above, there usually won't be coordinates $x_i$ such that $e^i_a = (\partial / \partial x_i)_a$, and so functions $g^{ij}$ as above wouldn't make sense. 
