= Solution
Work on a regular patch with $f(u)>0$. Differentiating the <surface of revolution> gives
$$
r_u=(f'\cos v,f'\sin v,h'),\qquad r_v=(-f\sin v,f\cos v,0).
$$
The arc-length condition gives $E=1$, $F=0$, $G=f^2$. Also $\|r_u\times r_v\|=f$, so take
$$
N=(-h'\cos v,-h'\sin v,f').
$$
The second derivatives yield
$$
e=f'h''-f''h',\qquad f_2=0,\qquad g_2=fh'.
$$
Thus the <first fundamental form> and <second fundamental form> are
$$
\boxed{\mathrm I=du^2+f^2dv^2,\qquad\mathrm{II}=(f'h''-f''h')du^2+fh'\,dv^2.}
$$
The opposite normal reverses the second displayed form. At a smooth pole the revolution coordinates degenerate, so <curvature> is understood by a regular local chart or by continuity.
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