Help Finding Orthonormal Frame and Coframe Based on First Fundamental Form Given the metric (for $x^2<1$)
$g=dx^2+2x dxdy+dy^2$ (in first fundamental form)
I'm trying to find the orthonormal frame and its coframe
I have found (I think) the orthonormal frame to be 
$e_1=\frac{\partial}{\partial x},e_2=\frac{\frac{\partial}{\partial y}-x\frac{\partial}{\partial x}}{\sqrt{1-x^2}}$
And I think the coframe is $e^1=dx,e^2=\sqrt{1-x^2}dy$, and each works for $e^ie_j=\delta^i_j$ except for the case $e^1e_2$ so I'm wondering if my orthogonal frame is right, and if so what am I missing to get the coframe
 A: This is too long for a comment and I would like to leave the OP with a chance to iron out some of the details . . . 
Given any non-zero vector field $e_{1}$, one can always apply the Gram-Schmidt Process to extend it to an orthonormal frame on its domain of definition.
Now, if you have a purported orthonormal frame and you want to verify this, you simply want to very that $g(e_i, e_j) = \delta_{ij}$.  Based on the question above, it appears that you realize this.  
In this case, it perhaps easiest to identify the metric $g$ with the matrix $\begin{pmatrix} 1 & x \\ x &1\end{pmatrix}$ and realizing that for any two vectors $\vec{u}, \vec{v}$, one has
$$
g(\vec{u}, \vec{v}) = \vec{u}^{t} \begin{pmatrix} 1 & x \\ x & 1\end{pmatrix} \vec{v},
$$
where we are identifying our vectors as column vectors with respect to the coordinate frame $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ and $\vec{u}^{t}$ denotes the transpose of the vector $\vec{u}$.
In this case, one has $e_{1} = \begin{pmatrix} 1 \\ 0\end{pmatrix}$ and $e_{2} = \begin{pmatrix} -\frac{x}{\sqrt{1 - x^2}}\\ \frac{1}{\sqrt{1 - x^2}}\end{pmatrix}$. 
In regards to the purported  orthonormal frame from the question, one should be able to check quickly and see that it is indeed an orthonormal frame.
Finally, concerning the matter of the coframe, this is mostly a matter of (point-wise) linear algebra and understanding how the change in the frame results in a corresponding change in the coframe.
In general, any two frames are related by an invertible matrix $A$ and you would like to know what the corresponding relationship between their respective coframes is.
In this instance, you have $\begin{pmatrix} e_{1} & e_{2} \end{pmatrix} = \begin{pmatrix} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} \end{pmatrix} A$, where $A = \begin{pmatrix} 1 & \frac{-x}{\sqrt{1 - x^2}}\\ 0 & \frac{1}{\sqrt{1 - x^2}}\end{pmatrix}$.  Note that this also gives 
$$
\begin{pmatrix} e_1 & e_2 \end{pmatrix} A^{-1}  = \begin{pmatrix} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} \end{pmatrix}.
$$
Your corresponding coframe $e^{1}, e^{2}$ can be expressed in terms of the coframe $dx, dy$ that is dual to the frame $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}$ as
\begin{align}
e^{1} &= \alpha dx + \beta dy\\
e^{2} &= \gamma dx + \delta dy,\\
\end{align}
or $\begin{pmatrix} e_1 \\ e_2 \end{pmatrix} = B \begin{pmatrix} dx \\ dy \end{pmatrix}$, where $B = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta\\\end{pmatrix}$ and $\alpha, \beta, \gamma, \delta$ are functions on the surface.
Charged with the task of identifying the components of the matrix $B$, one can simply evaluate the one forms $e^{1}$ and $e^{2}$ on the vectors of the original frame: $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$, where you will want to make use of the fact that you know $\frac{\partial}{\partial x}$ and ${\partial}{\partial y}$ in terms of $e_{1}$ and $e_{2}$ via the relationship $\begin{pmatrix} e_1 & e_2 \end{pmatrix} A^{-1}  = \begin{pmatrix} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} \end{pmatrix}$
For example, when one evaluates $e^{1} = \alpha dx + \beta dy$ on the vector $\frac{\partial}{\partial x}$, one will find that  $e^{1}\left(\frac{\partial}{\partial x}\right) = \alpha$ and that by expressing $\frac{\partial}{\partial x}$ in terms of the frame $e_1$ and $e_2$, one can express $\alpha$ in terms of the components of the matrix $A$ (or $A^{-1}$).
Paying attention to the individual computations, you will see that one is able to learn how to automatically change the coframe by using the same matrix (with the appropriate adjustments) that were used in the change of frame.
This ended up being a longer answer than I intended, but I hope that you find it helpful.
