Theorema Egregium (Gauss's Remarkable Theorem)

Work in coordinates where $F=0$ at the point of interest (an orthogonal parametrization, which always exists locally). Differentiating the identities $\mathbf{r}_u\cdot\mathbf{n}=0$ and $\mathbf{r}_v\cdot\mathbf{n}=0$ and using $e = \mathbf{r}_{uu}\cdot\mathbf{n}$, $g = \mathbf{r}_{vv}\cdot\mathbf{n}$ expresses $eg-f^2$ in terms of the second derivatives $\mathbf{r}_{uu}, \mathbf{r}_{uv}, \mathbf{r}_{vv}$ projected onto $\mathbf{n}$.

Because $\{\mathbf{r}_u, \mathbf{r}_v, \mathbf{n}\}$ form a basis of space at each point, every second derivative such as $\mathbf{r}_{uu}$ can be written as a combination of $\mathbf{r}_u$, $\mathbf{r}_v$ and $\mathbf{n}$, with the tangential coefficients being the Christoffel symbols — functions built only from $E, F, G$ and their derivatives with respect to $u$ and $v$, obtained by solving the linear system that comes from differentiating $E, F, G$.

Substituting these expressions into $\mathbf{r}_{uu}\cdot\mathbf{r}_{vv} - \mathbf{r}_{uv}\cdot\mathbf{r}_{uv}$ and comparing with $eg-f^2$ from the normal components, then simplifying with the compatibility (Gauss) equations, produces Brioschi's formula: $K$ expressed purely as a rational function of $E, F, G$ and their first and second partial derivatives with respect to $u$ and $v$, with no reference to $e, f, g$ left in the final expression.

Since an isometry between two surfaces is, by definition, a map that carries the first fundamental form of one surface exactly onto the first fundamental form of the other (the same $E, F, G$ as functions of the parameters), and Brioschi's formula computes $K$ from $E, F, G$ alone, the two surfaces must have equal Gaussian curvature at corresponding points. This proves the theorem.