My problem is to solve this exercise :
Let $S \subset \mathbb{R}^3$ be a sybmanifold of dimension $2$, and $x:U \mapsto S$ a local parametrization at $p = x(u,v) \in \mathcal{S}$. We have (admitted) $$ \langle (x_{vv}^T)_u (u,v) - (x_{vu}^T)_v (u,v) , x_u(u,v) \rangle = K \circ x(u,v) (EG-F^2)(u,v) $$ where $x_{vv}^T$ is the tangential component of $x_{vv}$. $K$ denotes the Gaus curvature and $x_u$ is the partial derivative in $u$.
Let $f$ be a local isometry bewteen $S$ and $S'$ two surfaces.
Prove that $K' \circ f = K$ where $K'$ is the Gauss curvature of $S'$.
I first proved that, if $x$ is a local parametrization of $S$ and $x'$ a local parametrization of $S'$ defined by $x'=f \circ x$ then :
$$\begin{align*} df_{x(u,v)} ([x_{uu}]^T (u,v)) &= [x'_{uu}]^T (u,v), \\ df_{x(u,v)} ([x_{vv}]^T (u,v)) &= [x'_{vv}]^T (u,v)\\ df_{x(u,v)} ([x_{uv}]^T (u,v)) &= [x'_{uv}]^T (u,v) \end{align*}$$
Then my idea was then to prove $K' \circ x' = K \circ x$ using the formula : $$ \langle (x_{vv}^T)_u (u,v) - (x_{vu}^T)_v (u,v) , x_u(u,v) \rangle = K \circ x(u,v) (EG-F^2)(u,v) $$
Since $E=E', F=F', G=G'$ thus it suffices to prove : $$\langle (x_{vv}^T)_u (u,v) - (x_{vu}^T)_v (u,v) , x_u(u,v) \rangle= \langle ({x'}_{vv}^T)_u (u,v) - ({x'}_{vu}^T)_v (u,v) , {x'}_u(u,v) \rangle$$
Since $\langle df_{x(u,v)} (X), df_{x(u,v)} (Y)\rangle = \langle X, Y \rangle $ I want to use this identity to prove the result.
$${x'}_u (u,v) = df_{x(u,v)} (x_u (u,v))$$ But this is not sufficient.
How can I finish ?
[I can explain some notations if you want, even if I think this is all standard]