 Solution

Take the parallel at height one,It has constant speed $\sqrt{2}$. Its distance from the axis is the constant $\rho =\sqrt{2}$, and its tangent is parallel to the parallel, so $\theta =0$ and it satisfies the Clairaut first integral for a surface of revolution.

However,whereas a normal vector to $x^2+y^2-z^2=1$ at $\gamma (t)$ isThese 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.