Let $\Gamma:\eta=(\eta_1(x_1,x_2),\eta_2(x_1,x_2),\eta_3(x_1,x_2))$ be a smooth surface in $\mathbb{R}^3$ with induced megetric $g=(g_{\alpha,\beta})$, where $g_{\alpha, \beta}=\partial _{\alpha}\eta\cdot \partial _{ \beta}\eta $. And let $ \sqrt{g}=\mathop{\mathrm{det}}\nolimits (g _{\alpha,\beta}) $. I would like to ask for help in proving \begin{gather} \displaystyle -\Delta _{g}\eta=\mathcal{H}n, \end{gather} where $ \mathcal{H} $ is twice the mean curvature of the surface $ \Gamma $, $ n:=\frac{\partial _{1}\eta \times \partial _{2}\eta}{|\partial _{1}\eta \times \partial _{2}\eta|} $ is the unit normal to $ \Gamma $, $ \Delta _{g} $ is the Laplace–Beltrami operator on $ \Gamma $ in terms of $ g $, which can be precisely given as follows \begin{gather} \displaystyle \Delta _{g} :=\sqrt{g}^{-1}\partial _{\alpha}[\sqrt{g} g ^{\alpha \beta}\partial _{\beta}],\ \ g ^{\alpha \beta}=(g _{\alpha \beta})^{-1}~(\mathrm{the\ inverse\ of\ the\ matrix}\ (g _{\alpha \beta})). \end{gather}
I hope someone would be kind enough to give me a hand.
Recall that the Hessian of a function $f$ on a Riemannian manifold $(M,g)$ is the covariant derivative of $df$ with respect to $\nabla$, the Levi-Civita connection of $g$. Explicitly, for $X,Y\in \mathscr{X}(M)$ we have $$\nabla^2f(X,Y)=X(df(Y))-df(\nabla_XY).$$ Here $f$ could be multi-valued.
In your case, $\eta$ is an embedding of a surface into $\mathbb{R}^3$ and in particular a multi-valued function on the surface. The differential $d\eta$ carries a vector tangent to the surface to its realization in $\mathbb{R}^3$. Let $X$ and $Y$ be vector fields on the surface. Note that covariant derivation in $\mathbb{R}^3$ is the usual differentiation. Hence, the above equation shows that, by definition, we have $$II(X,Y)=\nabla^2\eta(X,Y),$$where $II$ is my pour notation for the second fundamental form. In other words, the Hessian of $\eta$ coincides with the second fundamental form. To obtain the desired equality, take the traces of both sides.