= Solution
Write derivatives with respect to $u$. The arc-length condition is $f'^2+g'^2=1$. The tangent vectors are
$$
\sigma_u=(f'\cos v,f'\sin v,g'),\qquad \sigma_v=(-f\sin v,f\cos v,0).
$$
Their scalar products give the <first fundamental form>
$$
\boxed{I=du^2+f^2dv^2.}
$$
Choose the unit <normal vector> $N=(-g'\cos v,-g'\sin v,f')$, since $\sigma_u\times\sigma_v=fN$ and $f>0$. Scalar products of $N$ with the second <partial derivatives> give
$$
\boxed{II=(f'g''-g'f'')\,du^2+fg'\,dv^2.}
$$
The mixed coefficient is zero. Reversing $N$ reverses the <second fundamental form> but not the <Gaussian curvature>. Its formula is
$$
K=\frac{(f'g''-g'f'')fg'}{f^2}=\frac{f'g'g''-g'^2f''}{f}.
$$
Differentiating the arc-length identity gives $f'f''+g'g''=0$. Substitute it directly, without dividing by $g'$ (which may vanish), to obtain
$$
\boxed{K=-\frac{f''}{f}.}
$$
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