= Solution
With
$$
X(u,v)=(\sinh u\cos v,\sinh u\sin v,v),
$$
the coordinate tangent vectors satisfy
$$
X_u\cdot X_u=\cosh^2u,
\qquad X_u\cdot X_v=0,
\qquad X_v\cdot X_v=\cosh^2u.
$$
Thus the <first fundamental form> is
$$
\boxed{ds^2=\cosh^2u\,(du^2+dv^2)}.
$$
Geometrically, $S$ is the half of a <helicoid> with positive radius, winding upwards indefinitely.
Consider the upper half of the <catenoid>, parametrized as the <surface of revolution>
$$
Y(u,v)=(\cosh u\cos v,\cosh u\sin v,u),
\qquad u,v>0.
$$
It has the same first fundamental form, so $(u,v)\mapsto Y(u,v)$ defines a <local isometry>. By the invariance of <Gaussian curvature> under local isometry, the curvatures agree; directly, the conformal metric above gives
$$
\boxed{K=-\operatorname{sech}^4u}.
$$
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