Let $M$ be a $2$-dimensional Riemannian manifold, and ${\bf x}: U \subset \Bbb R^2 \to M$ be a parametrization of $M$. Suppose that $\bf x$ is orthogonal, that is, $F = \langle {\bf x}_u,{\bf x}_v\rangle = 0$. Let $\nabla$ be the Levi-Civita connection on $M$.
I have checked that $E_1 = \frac{{\bf x}_u}{\sqrt{E}}$ and $E_2 = \frac{{\bf x}_v}{\sqrt{G}}$ is an orthonormal frame on ${\bf x}(U)$, and that its dual forms are $\theta_1 = \sqrt{E} \ {\rm d}u$ and $\theta_2 = \sqrt{G} \ {\rm d}v$ (here, ${\rm d}u$ is dual to ${\bf x}_u$ and ${\rm d}v$ is dual to ${\bf x}_v$). Also, I computed the connection form $$\omega_{12} = -\frac{(\sqrt{E})_v}{\sqrt{G}} \ {\rm d}u + \frac{(\sqrt{G})_u}{\sqrt{E}} \ {\rm d}v.$$
Then, the exercise asks to check that: $$\omega_{12}({\bf x}_v) = \frac{1}{\sqrt{E}G}\langle \nabla_{{\bf x}_u}{\bf x}_v, {\bf x}_v\rangle = \frac{1}{\sqrt{E}G}\langle \nabla_{{\bf x}_v}{\bf x}_u, {\bf x}_v\rangle \quad \mbox{and} \\ \omega_{12}({\bf x}_u) = \frac{1}{\sqrt{E}G}\langle \nabla_{{\bf x}_v}{\bf x}_u, {\bf x}_u\rangle = \frac{1}{\sqrt{E}G}\langle \nabla_{{\bf x}_u}{\bf x}_v, {\bf x}_u\rangle.$$
I'm not sure this is right. I think that there's a typo, and should be $\sqrt{EG}$ instead of $\sqrt{E}G$ everywhere above. I'll show what I have done in this part:
We have $\omega_{12}({\bf x}_v) = \frac{(\sqrt{G})_u}{\sqrt{E}}$. Then: $$\frac{\partial}{\partial u}\sqrt{G} = \frac{1}{2\sqrt{G}}\frac{\partial}{\partial u}\langle {\bf x}_v, {\bf x}_v\rangle = \frac{1}{2\sqrt{G}} 2\langle \nabla_{{\bf x}_u}{\bf x}_v,{\bf x}_v\rangle = \frac{1}{\sqrt{G}}\langle \nabla_{{\bf x}_u}{\bf x}_v,{\bf x}_v\rangle.$$ This way, $\omega_{12}({\bf x}_v) = \frac{1}{\sqrt{EG}}\langle \nabla_{{\bf x}_u}{\bf x}_v, {\bf x}_v\rangle$ instead of what the question said. And $\omega_{12}({\bf x}_u)$ is analogous. Surely I could've made some silly mistake, but doesn't hurt to check.
Thanks!
It remains to show that $\omega_{12}( X_v)=\frac{1}{\sqrt{EG}}\left\langle\nabla_{ X_v} X_u, X_v\right\rangle$ and that $\omega_{21}( X_u)=\frac{1}{\sqrt{EG}}\left\langle\nabla_{ X_u} X_v, X_u\right\rangle$. I will show how to prove the first equality.
$$\omega_{12}(X_v)=\left\langle\nabla _{X_v}E_1, E_2\right\rangle=\left\langle\nabla _{X_v}\frac{X_u}{\sqrt E}, \frac{X_v}{\sqrt G}\right\rangle=\frac{1}{\sqrt G}\left\langle\frac{\nabla _{X_v}X_u}{\sqrt E}+X_v[\frac{1}{{\sqrt E}}]X_u, X_v\right\rangle=\frac{1}{\sqrt {EG}}\left\langle\nabla _{X_v}X_u, X_v\right\rangle $$
As we wished. And you are right: the person who wrote the question made a typo. Also, the second sequence of equalities is wrong. Change $\omega_{12}$ to $\omega_{21}$.