Let $S$ be a set of points $ (x,y,z)\in\mathbb{R^3}$ that satisfy the equation $x^3+y^2+z^2=1$, $p_1=(0,1,0)$ and $p_2=(0,0,1)$ points in $S$ with the same gaussian curvature,prove that $f: S \to S$ defined by $f(x, y, z) = (x, z, y)$ is an isometry that swaps $p_1$ and $p_2$.
is this demonstration corrrect?
We have that $f$ is a diffeomorphism that waps $p_1$ and $p_2$, to demostrate that $f$ is a isometry we need to mostrate that it preserves the first fundamental form, i.e. $<df_p[w_1],df_p[w_2]>=<w_1,w_2>$ for all $w_1,w_2 \in T_pS$ and $p \in S$. But chosen $w=(x,y,z) \in T_pS$ we have $df_p[w]=(x,z,y)$, so $<df_p[w_1],df_p[w_2]>=x_1x_2+z_1z_2+y_1y_2=x_1x_2+y_1y_2+z_1z_2=<w_1,w_2>$, so $f$ is an isometry.
An other my question is, if this demonstration is correct so let's now consider the same surface $S$ that I defined. Take $p_3=(1,0,0)$. We observe that $p_3$ and $p_1$ have different Gaussian curvatures. Now, if we consider the function $g:S->S$ defined as $g(x,y,z)=(z,y,x)$ and apply the same procedure I used to show that $f$ is an isometry to $g$, what goes wrong in this case? (knowing that isometry preserve gaussian curvature so in this case it can 't exist a isometry )