= Solution
For symmetric quadratic coefficient matrices, the <Euler-Lagrange equations> give $T\ddot q+Vq=0$, where $T$ is the positive-definite <mass matrix>. Substitute a normal-mode oscillation $q=v\cos(\omega t+\phi)$ to get
$$
\boxed{(V-\omega^2T)v=0,\qquad\det(V-\omega^2T)=0.}
$$
The positive roots $\omega^2$ determine the squared normal frequencies and their nonzero null vectors give mode shapes. Repeated eigenvalues give multidimensional mode spaces.
The particle's <kinetic energy> is $m|\dot q|^2/2$. For a taut elastic string at length $r=\sqrt{x^2+y^2+z^2}$, <Hooke's law> gives tension $\lambda(r-a)/a$ and elastic energy $\lambda(r-a)^2/(2a)$. With $\lambda=mg$ and gravitational potential $mgz$, this gives exactly
$$
T=\frac m2(\dot x^2+\dot y^2+\dot z^2),\qquad
V=mgz+\frac{mg}{2a}(r-a)^2.
$$
This is the taut-string regime containing the equilibrium. Differentiating the potential gives
$$
\boxed{\ddot x=-\frac ga\frac{r-a}{r}x,\quad
\ddot y=-\frac ga\frac{r-a}{r}y,\quad
\ddot z=-g-\frac ga\frac{r-a}{r}z.}
$$
At $(0,0,-2a)$, $r=2a$ and all three accelerations vanish.
With $z=-2a+\zeta$, expansion to quadratic order gives $r=2a-\zeta+(x^2+y^2)/(4a)+O(3)$. Substitution cancels the linear terms, leaving
$$
\boxed{V=-\frac32mga+\frac{mg}{4a}(x^2+y^2)+\frac{mg}{2a}\zeta^2+O(3).}
$$
The <mass matrix> remains $mI$, and the <stiffness matrix> is $\operatorname{diag}(mg/(2a),mg/(2a),mg/a)$. Therefore
$$
\boxed{\omega_x=\omega_y=\sqrt{\frac g{2a}},\qquad
\omega_z=\sqrt{\frac ga}.}
$$
The horizontal frequency is doubly degenerate: a linearly polarized oscillation in any horizontal direction is a <normal mode>. The remaining mode is vertical stretching and contraction of the string, with horizontal coordinates zero.
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