Let $M \subset \mathbb{R}^3$ be an embedded surface, then $II(X, Y)=\langle X, AY \rangle$, where $II$ is the second fundamental form, and $A:T_xM \to T_xM$ is an endomorphism. If all points of $M$ are umbilical cords, ie the eigenvalues of $A$ coincide for every $x \in M$, then the principal curvatures, that is the eigenvalues of $A$, are constant.
Here's what I got:
Let $v \in T_xM$ be an eigenvector of $A$, with eigenvalue $\lambda$. Then there's a geodesic $\gamma: (-\varepsilon,\varepsilon) \to M$ with $\gamma(0)=x, \ \gamma'(0)=v.$ Let $X$ be a section extending the geodesic, then: $$II(X,X):=\langle \nabla_X X, \ n \rangle=\langle \nabla_{\gamma'} \gamma', \ n \rangle=\langle 0, \ n \rangle=0$$
But we also have $$II(X,X)=\langle X, \ AX \rangle=\lambda \langle v, v \rangle= \lambda ||v||^2 $$
Hence $\lambda ||v||^2=0 \Rightarrow \lambda = 0.$ Hence the eigenvalues of $A$ are constant.
I'm not convinced my proof is correct so some feedback would be great.