I have a fundamental misunderstanding of one of the central ideas in any introduction to differential geometry of surfaces – that isometries of surfaces preserve the coefficients of the so-called "first fundamental form" of a surface (hence the importance of this form). As far as I understand, isometries by definition preserve the length element on the surface – that is, given an intrinsic parametrization $r(u,v)$ of a surface, the length element before isommetry is $ds^2=Edu^2+2Fdudv+Gdv^2$, and the length element after isommetry is $ds'^2=E'du'^2+2F'du'dv'+G'dv'^2$ and both length elements are equal, $ds^2=ds'^2$.
But how can one deduce from the equality of length elements the equality of the coefficients of both forms (that $E=E',F=F',G=G'$)? in all descriptions of the implications of the "Theorema Egregium" there is a conclusion that if the Gaussian curvature depends solely on the coefficients of the first fundamental form and their first and second derivatives than it is preserved under isommetries, so there is a "tacit" assumption here that isommetries preserve the coefficients of first fundamental form.
I know this is probably a very basic question for anyone who is familiar with the classical differential geometry of surfaces, so any useful comment will be appreciated.
Best Answer
Ted Shifrin has made an important caveat, and I posted a link in a comment. But rereading your question, I think the issue is not mainly about differential geometry at all. Rather it’s algebraic.
As you indicate in your first paragraph, you have two polynomials in two variables (both of degree 2, but that doesn’t matter) that are equal for all values of the variables in an open neighborhood of the origin: $$Ex^2+2Fxy+Gy^2=E’x^2+2F’xy+G’y^2$$ That $x$ and $y$ are differentials doesn’t matter.
As you probably know already, if two polynomials in one variable agree for infinitely many values— $f(x)=g(x)$ for infinitely many $x$’s— then $f\equiv g$.
If $f(x,y)=g(x,y)$ for all $(x,y)$ in a neighborhood of the origin, then write $f(x,y)=\sum_i f_i(x)y^i$ where the $f_i$’s are polynomials in $x$. Likewise for $g$. For any fixed value of $x$ we have polynomials in $y$ that agree for infinitely many $y$’s, so $f_i(x)=g_i(x)$ by the result just mentioned. But this is true for infinitely many $x$’s, so we must have $f_i\equiv g_i$ for all $i$.