= Solution
For the <surface of revolution> $r(\eta,\zeta)=(\eta,f(\eta)\cos\zeta,f(\eta)\sin\zeta)$, the coordinate tangent <vectors> have squared lengths $1+f'^2$ and $f^2$, and their dot product is zero. Hence its <first fundamental form> is
$$
\boxed{ds^2=A(\eta)d\eta^2+f(\eta)^2d\zeta^2,\qquad A=1+f'^2.}
$$
Use the energy <Lagrangian> $L=A\dot\eta^2+f^2\dot\zeta^2$. Its Euler-Lagrange equations are
$$
2A\ddot\eta+A'\dot\eta^2-2ff'\dot\zeta^2=0,\qquad
\frac d{dt}(2f^2\dot\zeta)=0.
$$
For a meridian $\zeta=\zeta_0$, choose arc length as parameter, so $\dot\eta=1/\sqrt A$. Differentiating gives $2A\ddot\eta+A'\dot\eta^2=0$, while the second equation is automatic. Part (b) makes this path an extremal of length. Thus \b[every meridian is a <geodesic>].
For a nonconstant parallel $\eta=\eta_0$, choose constant nonzero $\dot\zeta$. The second equation holds, and the first reduces to $f(\eta_0)f'(\eta_0)\dot\zeta^2=0$. Since $f>0$, this is equivalent to
$$
\boxed{f'(\eta_0)=0.}
$$
Conversely that condition makes both equations hold at constant speed, hence supplies a length extremal. Geometrically, a parallel is geodesic exactly at a stationary radius of the surface.
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