= Solution
The <Theorema Egregium> states that the <Gaussian curvature> of a surface in $\mathbb R^3$ is determined entirely by its <first fundamental form>; in particular a local <isometry> preserves it.
Let $r(u^1,u^2)$ be a regular parametrization, $g_{ij}=r_i\cdot r_j$, $n$ a unit normal, and $b_{ij}=r_{ij}\cdot n$ the <second fundamental form>. Differentiating the metric and decomposing $r_{ij}$ into tangent and normal parts gives the Gauss formula
$$
r_{ij}=\Gamma^k_{ij}r_k+b_{ij}n,\qquad
\Gamma^k_{ij}=\frac12g^{k\ell}(\partial_i g_{j\ell}+\partial_jg_{i\ell}-\partial_\ell g_{ij}).
$$
Differentiating $n\cdot r_j=0$ gives $n_i=-b_i{}^kr_k$, the Weingarten formula. Compute two successive ambient derivatives of a tangent vector and subtract them. Ambient derivatives commute, so their tangential parts yield the <Gauss equation>
$$
R(X,Y)Z=b(Y,Z)SX-b(X,Z)SY,
$$
where $S$ is the shape operator and $R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z$ is the intrinsic curvature of the connection with the displayed Christoffel symbols. Taking $X=\partial_1,Y=\partial_2,Z=\partial_2$ and the inner product with $\partial_1$ gives
$$
\langle R(\partial_1,\partial_2)\partial_2,\partial_1\rangle=b_{11}b_{22}-b_{12}^2.
$$
But $K=\det S=\det(b)/\det(g)$, so
$$
\boxed{K=\frac{\langle R(\partial_1,\partial_2)\partial_2,\partial_1\rangle}{g_{11}g_{22}-g_{12}^2}.}
$$
Both the connection and its curvature on the right are computed from $g$ and its derivatives alone. This proves the theorem, rather than merely using its name as a substitute for the argument.
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