= Solution
For a circular mid-plane orbit, radial force balance in the <Newtonian gravitational potential> is $r\Omega^2=\Phi_{,r}$. Hence
$$
\boxed{\Omega^2(r)=\frac1r\Phi_{,r}(r,0)}.
$$
Axisymmetry conserves the <specific angular momentum> $j=r^2\dot\phi$. At fixed $j$, radial motion obeys $\ddot r=-\partial_r\Phi_{\rm eff}$, where the <effective potential> is $\Phi_{\rm eff}=\Phi+j^2/(2r^2)$. Linearize around a circular radius $r_0$, writing $r=r_0+\xi$. Since $j^2=r_0^3\Phi_{,r}(r_0,0)$ there,
$$
\ddot\xi+\kappa^2\xi=0,\qquad
\boxed{\kappa^2=\Phi_{,rr}(r_0,0)+\frac3{r_0}\Phi_{,r}(r_0,0)}.
$$
This is the <radial epicyclic frequency>; equivalently $\kappa^2=r^{-3}d(r^4\Omega^2)/dr$.
Reflection symmetry gives $\Phi_{,z}(r,0)=0$ and $\Phi_{,rz}(r,0)=0$, so vertical and radial displacements decouple at linear order. The vertical equation is $\ddot z=-\Phi_{,zz}(r_0,0)z$, giving the <vertical epicyclic frequency>
$$
\boxed{\Omega_z^2(r_0)=\Phi_{,zz}(r_0,0)}.
$$
Positive squared frequencies give oscillatory stability; a negative squared frequency gives exponential instability rather than a real oscillation frequency.
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