= Solution
Choose $f(u)=\sin u$ and $h(u)=\cos u$ for $0\leq u\leq\pi$. Adding the smooth poles at the endpoints gives the unit <sphere>. Its meridian is parametrized by <arc length>, and $-f''/f=1$ away from the poles, hence everywhere by continuity. Thus \b[the unit <sphere> has constant <Gaussian curvature> one].
There is \b[no <compact> smooth <surface of revolution> without boundary of constant <Gaussian curvature> minus one] in this setting. On a <compact> such <surface>, the distance $f$ from the rotation axis attains a positive maximum. The maximum lies on a regular revolution patch, where the meridian <arc length> coordinate gives $f''\leq0$. Hence $K=-f''/f\geq0$ there, contradicting $K=-1$. Equivalently, constant negative <curvature> would require $f''=f>0$, excluding any positive radial maximum. Compact patches with boundary would not be excluded by this argument, but they are not closed surfaces.
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