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For a local parametrization $\varphi (u,v)$, write the first fundamental form and second fundamental form asThe Gaussian curvature is

To prove the Theorema Egregium, set $g_{ij}=⟨{\varphi }_i,{\varphi }_j⟩$ and decomposeDifferentiating $g_{ij}$ and solving for the tangential coefficients givesso every Christoffel symbol depends only on $E,F,G$ and their first derivatives. Equality of mixed third derivatives yields the Gauss equationThe Riemann curvature tensor is formed from first derivatives and quadratic products of the Christoffel symbols, and hence from the first and second derivatives of $E,F,G$. Consequentlydepends only on the first fundamental form.

For a chosen unit normal field $N$, the second fundamental form isIn a local parametrization $\varphi$, its coefficients, up to the equivalent simultaneous sign convention, areThe Gaussian curvature is the determinant of the shape operator, henceTheorema Egregium states that $K$ is determined entirely by the first fundamental form. In particular, it is intrinsic and is preserved by every local isometry.

The Gaussian curvature $K$ is the determinant of the shape operator, equivalently the product of the two principal curvatures. Yes,This is the Theorema Egregium: Gaussian curvature depends only on the first fundamental form, which a local isometry preserves.

Gauss's Theorema Egregium expresses Gaussian curvature intrinsically from the first fundamental form, hence $K\circ \varphi =K$. Merely preserving curvature level sets is not sufficient: when $K$ is constant, every diffeomorphism does so, though most are not isometries.