= Solution
Apply the <dispersion relation> separately to the two components and subtract:
$$
m(\omega_0+\delta\omega)=(k_0+\delta k)^2,\quad
m(\omega_0-\delta\omega)=(k_0-\delta k)^2
\quad\Longrightarrow\quad
\frac{\delta\omega}{\delta k}=\frac{2k_0}{m}.
$$
In the narrow-band limit this is the <group velocity>, since $\omega(k)=k^2/m$ gives
$$
\boxed{v_g=\frac{d\omega}{dk}=\frac{2k}{m}=2v_p.}
$$
Equivalently on the positive branch $k=\sqrt{m\omega}$, $dk/d\omega=m/(2k)$ and its reciprocal is $2k/m$. The factor of two applies to signed velocities on either nonzero propagation branch.
There is a small but useful finite-band qualification. Adding the two component relations gives $m\omega_0=k_0^2+(\delta k)^2$. Thus the carrier formed by their exact midpoints has <phase velocity> $\omega_0/k_0=k_0/m+(\delta k)^2/(mk_0)$, and its finite-band envelope-to-carrier velocity ratio is
$$
\frac{\delta\omega/\delta k}{\omega_0/k_0}
=\frac{2k_0^2}{k_0^2+(\delta k)^2}.
$$
The exact derivative relation $v_g=2v_p$ refers to a single central point on the <dispersion relation>; the two-tone midpoint ratio approaches two as $\delta k/k_0\to0$.
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