= Solution
Take the parallel at height one,
$$
\gamma(t)=(\sqrt2\cos t,\sqrt2\sin t,1).
$$
It has constant speed $\sqrt2$. Its distance from the axis is the constant $\rho=\sqrt2$, and its tangent is parallel to the parallel, so $\theta=0$ and it satisfies the <Clairaut first integral for a surface of revolution>.
However,
$$
\ddot\gamma=(-\sqrt2\cos t,-\sqrt2\sin t,0),
$$
whereas a <normal vector> to $x^2+y^2-z^2=1$ at $\gamma(t)$ is
$$
(\sqrt2\cos t,\sqrt2\sin t,-1).
$$
These <vectors> are not parallel. By the <ambient acceleration criterion for a surface geodesic>, $\gamma$ is not a geodesic. Thus constant speed together with Clairaut's relation is not sufficient for the geodesic equations.
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