= Solution
\b[No.] Fix $0<A<1$ and, near $s=0$, let
$$
f(s)=A\cos s,
\qquad
g(s)=\int_0^s\sqrt{1-A^2\sin^2t}\,dt.
$$
Then $f'^2+g'^2=1$, so this is a regular unit-speed profile with $f>0$ on a sufficiently small interval. Rotate it around the $z$-axis. For an arc-length <surface of revolution>, $K=-f''/f$, and therefore
$$
K=-\frac{-A\cos s}{A\cos s}=1.
$$
At $s=0$, the formulas from part (a) give the <principal curvature>[principal curvatures]
$$
k_1=\frac1A,
\qquad k_2=A.
$$
They are unequal because $A\ne1$, so the point is not an <umbilical point>. Every point of a sphere is umbilical. Consequently no neighbourhood of this point can lie on a sphere, giving a <non-spherical surface of constant Gaussian curvature one>.
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