Differential geometry – Double frame in Riemannian metrics.

Suppose we have a Riemannian metric. $ ds ^ 2 = Edu ^ 2 + 2Fdudv + Gdv ^ 2 $ in a coordinated local neighborhood $ (U; (u, v)) $ Prove that for the following vector fields:

$$ e_ {1} = frac {1} { sqrt {E}} frac { partial} { partial u}, e_ {2} = frac {-1} { sqrt {EG- F ^ 2}} left ( frac {F} { sqrt {E}} frac { partial} { partial} – sqrt {E} frac { partial} { partial v} right ) $$

the $ 1- $shapes:
$$ omega_1 = sqrt {E} left (du + frac {F} {E} dv right), omega_2 = sqrt { frac {EG-F ^ 2} {E}} dv $ PS
satisfy
$$ omega_i (e_k) = delta_ {ik} $$

My work: Let $ p (u, v) $ A differentiable function, I believe that I must show fervently that $ omega_1 (e_1 (p)) = p $, that is to say $ omega_1 (e_1) = 1 $ The identity function.

$$ omega_1 left ( tfrac {1} { sqrt {E}} tfrac { partial} { partial u} (p) right) = tfrac {1} { sqrt {E}} tfrac { partial} { partial u} ( omega_1 (p)) $$
I do this because a $ 1-form $ $ alpha $ It is such that $ alpha (fX) = f alpha (X) $. Then it is correct that
$ omega_1 (p) = sqrt {E} left (pdu + frac {F} {E} pdv right)? $
and then apply the partial derivative, my problem is that I do not know how to operate the $ 1 $-to form. Anyone can guide me on how I can achieve the result or an explicit way to operate $ omega_1 $ Y $ omega_2 $?

I am using the book "Umehara, differential geometry of surfaces".