= Solution
For an oriented <smooth surface> $S$, the <Gauss map> sends $p$ to the chosen unit <normal vector> $N(p)\in S^2$. Since $|N|^2=1$, differentiation shows that $DN_p(X)$ is perpendicular to $N(p)$ and hence lies in $T_pS$.
In a local parametrization $\phi(u,v)$ with $n=N\circ\phi$, differentiating
$$
n\mathbin{\cdot}\phi_u=n\mathbin{\cdot}\phi_v=0
$$
gives
$$
n_u\mathbin{\cdot}\phi_v=-n\mathbin{\cdot}\phi_{uv}
=n_v\mathbin{\cdot}\phi_u.
$$
Thus the bilinear form $(X,Y)\mapsto DN_p(X)\mathbin{\cdot}Y$ is symmetric, so $DN_p$ is <self-adjoint operator>[self-adjoint]. The <Gaussian curvature> is
$$
\kappa=\det(DN_p).
$$
Writing the coefficients of the <first fundamental form> as
$$
E=\phi_u^2,\qquad F=\phi_u\mathbin{\cdot}\phi_v,\qquad G=\phi_v^2
$$
and those of the <second fundamental form> as
$$
e=n\mathbin{\cdot}\phi_{uu},\qquad
f=n\mathbin{\cdot}\phi_{uv},\qquad
g=n\mathbin{\cdot}\phi_{vv},
$$
one obtains
$$
\boxed{\kappa=\frac{eg-f^2}{EG-F^2}}.
$$
At an <umbilical point>[umbilic point], the self-adjoint map $DN_p$ has a repeated eigenvalue, so it is a scalar map. If every point is umbilic, there is a function $\lambda$ with
$$
n_u=\lambda\phi_u,\qquad n_v=\lambda\phi_v.
$$
Equality of mixed partial derivatives gives
$$
\lambda_v\phi_u=\lambda_u\phi_v.
$$
The two tangent vectors are linearly independent, hence $\lambda_u=\lambda_v=0$. Since $\mathbb R^2$ is connected, $\lambda$ is constant.
If $\lambda=0$, then $n$ is constant and
$$
\partial_u(n\mathbin{\cdot}\phi)
=\partial_v(n\mathbin{\cdot}\phi)=0,
$$
so $S$ lies in a plane. If $\lambda\ne0$, then
$$
\partial_u(n-\lambda\phi)
=\partial_v(n-\lambda\phi)=0.
$$
Thus $n-\lambda\phi=c$ is constant, and
$$
\left|\phi+\frac c\lambda\right|
=\frac{|n|}{|\lambda|}
=\frac1{|\lambda|}.
$$
\b[Therefore $S$ lies in a sphere of radius $1/|\lambda|$. This proves that the surface is part of a plane or part of a sphere.]
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