I'm trying to show that the two-dimensional elliptic paraboloid $S$, i.e. the graph of the function $f : \mathbb R^2 \to \mathbb R$ given by $f(x,y) = x^2 + y^2$, is isotropic at only one point (namely its vertex). I've reduced the problem to showing that $S$ is not isotropic at the point $p := (x,0,x^2)$ for any $x \in \mathbb R$, but I'm having trouble doing this.
I know the shape operator has orthogonal eigenvectors given by $V_1 = x\partial_x + y\partial_y + 2(x^2+y^2)\partial_z$ and $V_2 = -y\partial_x + x\partial_y$, and the respective eigenvalues (i.e. principle curvatures) are $\kappa_1 = 2r^{-3}$ and $\kappa_2 = 2r^{-1}$, where $r := \sqrt{1+4(x^2+y^2)}$. Looking at the point $p = (x,0,x^2)$ and normalizing, these simplify to $U_1 = \frac{1}{\sqrt{1+4x^2}}\left(\partial_x + 2x\partial_z\right)$ and $U_2 = \partial_y$.
What I'm trying: I think this can be solved using the shape operator, second fundamental form, or the Theorema Egregium, but I'm not sure how. Ultimately I want to find an orthonormal basis $\{E_1, E_2\}$ of $T_p S$ that cannot be attained from an isometry of $S$ fixing $p$. I've been trying to use linear algebra with the shape operator and second fundamental form to do this, but I keep going in circles. Can anyone give me a hint?
Firstly we want something intrinsic to the paraboloid so it makes sense to look at how the Gaussian curvature varies point to point. By the expressions you have provided the Gaussian curvature is maximized at the unique point $(0,0)$.
The idea will be to find an isometry that does not fix $(0,0)$. Pick a $v \in T_{(0,0)}M$ in a normal neighbourhood about $(0,0)$. Let $p = \exp_{(0,0)}(v/2)$ and $q = \exp_{(0,0)}(v)$. Now work in the tangent space at $p$ where we let $w \in T_pM$ be defined by $(0,0) = \exp_p(w)$ and $q = \exp_p(-w)$. If $M$ is isotropic at $p$ then there exists an isometry $\varphi$ with $\varphi_* w = -w$. It follows that $\varphi((0,0)) = q \neq (0,0)$ as desired.
Now for the easier part, if $b_1,b_2$ is an orthonormal basis at $(0,0)$ then by the Theorem Egregium and the fact that $\varphi$ is an isometry $$K((0,0)) = Rm(b_1,b_2,b_2,b_1)_{(0,0)} = Rm(\varphi_*b_1,\varphi_*b_2,\varphi_*b_2,\varphi_*b_2)_{q} = K(q),$$ contradicting the fact that the Gaussian curvature, $K(\cdot)$, is attains its unique maximum at $(0,0)$.