With
$$
p = (x,y,z)=\left(a \sin (\phi (\tau )) \cos (t(\tau )),a \sin (\phi (\tau )) \sin (t(\tau )),a \left(\cos (\phi (\tau
))+\ln \left(\tan \left(\frac{\phi (\tau )}{2}\right)\right)\right)\right)
$$
we have
$$
v = \dot p = \left(a \left(\phi '(\tau ) \cos (\phi (\tau )) \cos (t(\tau ))-\sin (\phi (\tau )) t'(\tau ) \sin (t(\tau
))\right),a \left(\phi '(\tau ) \cos (\phi (\tau )) \sin (t(\tau ))+\sin (\phi (\tau )) t'(\tau ) \cos (t(\tau
))\right),a \phi '(\tau ) \cos (\phi (\tau )) \cot (\phi (\tau ))\right)
$$
so the Lagrangian reads
$$
L = \frac 12 m v\cdot v - m g z(\phi(\tau)) = \frac{1}{2} a m \left(a \phi '(\tau )^2 \cot ^2(\phi (\tau ))+a \sin ^2(\phi (\tau )) t'(\tau )^2-2 g \left(\cos
(\phi (\tau ))+\ln \left(\tan \left(\frac{\phi (\tau )}{2}\right)\right)\right)\right)
$$
the movement equations
$$
\left\{
\begin{array}{rcl}
\phi ''(\tau )& = & \sin (\phi (\tau )) \left(\sin (\phi (\tau )) \tan (\phi (\tau )) t'(\tau
)^2-\frac{g}{a}\right)+\phi '(\tau )^2 \csc (\phi (\tau )) \sec (\phi (\tau )) \\
t''(\tau )& = & -2 \phi '(\tau ) \cot (\phi (\tau )) t'(\tau ) \\
\end{array}
\right.
$$
Solving those DEs for $a = 1, m = 1, g = 10, \phi(0) = 0.5, t(0) = 0.5, \dot\phi(0) = 1,\dot t(0)= 2$ we obtain the corresponding orbit in blue, shown in the attached plot
Follows the orbit projected into the $XY$ plane
Follows the plot for $z$ (black), $t(\tau)/10$ (red), and $r = \sqrt{x^2+y^2}$ (blue)
