Gauss-Codazzi Equations of embedded surface in $\mathbb{R}^3$. In this paper, the authors derive the Gauss-Codazzi equations for a surface embedded in $\mathbb{R}^3$, but I cannot reproduce it. It's a classical result, but I want to make sure I'm not missing something. I'll reproduce everything that's needed (I think!) here.
For the embedding $X:\Sigma \to\mathbb{R}^3$, we can choose as basis vectors on the embedded surface $\{e_{(i)}\}=(X_z,X_{\bar{z}},N)$, where $X_z=\partial_z X$ and $N$ is the normal vector to the surface. These vectors satisfy
$$X_z\cdot X_{\bar{z}}=\frac{f^2}{2}\tag{1}$$
$$X_z\cdot X_z=0\tag{2}$$
$$X_{\bar{z}}\cdot X_{\bar{z}}=0.\tag{3}$$
We also have definitions for the Hopf differential $Q$
$$X_{zz}\cdot N=Q,\tag{4}$$
$$X_{\bar{z}\bar{z}}\cdot N=Q^*\tag{5}$$
and the mean curvature $H$
$$X_{\bar{z}z}\cdot N=\frac{1}{2}f^2H.\tag{6}$$
Now the statement in the paper is "Using the condition for the commutivity of the derivatives $e_{(i),z\bar{z}}=e_{(i),\bar{z}z}$ we can derive the Gauss-Codazzi equations:"
$$ff_{z\bar{z}}-f_{z}f_{\bar{z}}+\frac{1}{4}f^4H^2=|Q|^2,\qquad Q_{\bar{z}}=\frac{1}{2}f^2H_z,\qquad Q^*_{z}=\frac{1}{2}f^2H_\bar{z}.$$
We don't have explicit forms for the second derivatives of the basis vectors, so the only idea I had was to take derivatives of those various inner products until we have the various derivatives of the unit vectors, set the ones equal to each other that should be, and that should be that. For instance,
$$\partial_{\bar{z}}(X_{zz}\cdot N)=X_{zz\bar{z}}\cdot N+X_{zz}\cdot N_\bar{z}=Q_{\bar{z}}\tag{7}$$
$$\partial_z(X_{z\bar{z}}\cdot N)=X_{z\bar{z}z} \cdot N+X_{z\bar{z}}\cdot N_z=ff_zH+\frac{1}{2}f^2H_z\tag{8}$$
If I subtract those, the first terms should cancel and I will get
$$X_{zz}\cdot N_\bar{z}-X_{z\bar{z}}\cdot N_z=Q_\bar{z}-ff_zH-\frac{1}{2}f^2H_z\tag{9}$$
Now if I want something like the first term, I can take two derivatives of the orthogonality condition:
$$\partial_{\bar{z}z}(X_{z}\cdot N)=0=X_{zz\bar{z}}\cdot N+X_{zz}\cdot N_\bar{z}+X_{z\bar{z}}\cdot N_{z}+X_z\cdot N_{z\bar{z}}\tag{10}$$
But that gives me back another factor with three derivatives - in fact, the term I already got rid of. So now this is feeling like I'm going in a circle. Does anyone have any insight into the specific pattern of identities which might get me to the right answer? Or, if there is a simple fact I am missing?
(Notice that I've numbered these equations, so you can simply give instructions rather then typing them all out if you see the solution. Also, demonstrating how to get just one of the required equations will be sufficient, I'm sure I will see how to get the others based on your solution.)
 A: $\newcommand{\barz}{\overline{z}}$I'll try to help you - what kind of math student would I be if I didn't solve other people's problems instead of my own to procrastinate? Put on some good music, because this will be long.
If $V$ is any vector, writing $V = aX_z+bX_\barz+cN$, dotting it with everything and solving for the coefficients, we have $$V = \frac{2}{f^2}\left( (V\cdot X_\barz)X_z+(V \cdot X_z)X_\barz\right)+(V\cdot N)N.$$
In particular, differentiating $X_z \cdot X_z = X_\barz \cdot X_\barz=0$ w.r.t. to $z$ and $\barz$ we obtain these nice relations: $$X_{zz}\cdot X_z = X_{z\barz}\cdot X_\barz = X_{z\barz}\cdot X_z = X_{\barz\barz}\cdot X_\barz = 0.$$Differentiating $X_z \cdot X_\barz = f^2/2$ w.r.t. to $z$ and $\barz$ and using the above we get $$X_{zz}\cdot X_\barz = ff_z, \quad X_{\barz\barz}\cdot X_z = ff_\barz.$$
Since we're at it, differentiating $N \cdot X_z = N\cdot X_\barz = 0$ in all ways possible, and using the definition of the Hopf differential, we get $$N_z \cdot X_z = -Q, \quad N_\barz \cdot X_\barz = -Q^\ast, \quad N_\barz \cdot X_z = N_z \cdot X_\barz = -\frac{f^2}{2}H.$$
Now, the author says in the paper to look at $\vec{e}_{(i),z\barz} = \vec{e}_{(i),\barz z}$, right? For $i=1$ this means $X_{zz\barz}= X_{z\barz z}$. Our expression for $V = X_{z\barz}$ is $$X_{z\barz} = \frac{f^2}{2}HN,$$and attacking that with $\partial_z$ gives
\begin{align}
X_{z\barz z}&= ff_zHN + \frac{f^2}{2}H_zN + \frac{f^2}{2}HN_z \\ &= \frac{f^2}{2}H \left( \frac{2}{f^2}\left( -\frac{f^2}{2}HX_z-QX_\barz)\right) \right)+\left(ff_zH+\frac{f^2}{2}H_z \right)N \\ &= -\frac{f^2}{2}H^2 X_z - QH X_\barz+\left(ff_zH+\frac{f^2}{2}H_z \right)N .
\end{align}
Note that I used $V = N_z$ here. Now we pick $V = X_{zz}$: $$X_{zz}= \frac{2f_z}{f}X_z + QN,$$and shoot at it with $\partial_\barz$:
\begin{align}
X_{zz\barz} &= \left( \frac{2f_{z\barz}f-2f_zf_\barz}{f^2}\right)X_z+\frac{2f_z}{f}X_{z\barz} + Q_\barz N + Q N_\barz \\ &= \left( \frac{2f_{z\barz}f-2f_zf_\barz}{f^2}\right)X_z +Q\frac{2}{f^2}\left(-Q^\ast X_z-\frac{f^2}{H}X_\barz \right)+ (ff_zH+Q_\barz)N \\ &= \frac{2}{f^2}\left(f_{z\barz}f-f_zf_\barz - |Q|^2 \right)X_z - QHX_\barz +(ff_zH+Q_\barz)N .
\end{align}
Comparing the coefficient of $X_z$ in both third derivatives gives equation $\rm (19)$ in the paper - your first Gauss-Codazzi equation. Comparing the coefficient  of $N$ gives equation  $\rm (20)$ - your second equation. To get the remaining one you'd probably have to repeat this strategy looking at $\vec{e}_{(i),z\barz} = \vec{e}_{(i),\barz z}$ for $i=2$, that is, $X_{\barz z \barz} = X_{\barz \barz z}$.
