= Solution
On a branch where $(\rho,\phi)$ are valid <local coordinates>, the <inclusion map> is $j(\rho,\phi)=(x(\rho),\rho\cos\phi,\rho\sin\phi)$, with $x^2=1-a^2\rho^2$. Its <differential> gives
$$
\boxed{j_*\partial_\phi=-z\partial_y+y\partial_z=:K.}
$$
Strictly this is first a <vector field along a map> on the surface. The displayed formula extends it canonically to a <smooth> <vector field> on all of $\mathbb R^3$. Its <integral curves of a vector field> through $(x_0,y_0,z_0)$ obey $\dot x=0$, $\dot y=-z$, $\dot z=y$, hence
$$
\boxed{x(s)=x_0,\qquad y(s)+iz(s)=(y_0+iz_0)e^{is}.}
$$
These are circles around the $x$-axis, traversed with angular speed one; points on the axis are stationary. Since $K(x^2+a^2(y^2+z^2))=0$, curves starting on the ellipsoid stay there. The local angular chart can expire while the ambient <flow map> and the geometric curve remain well defined.
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