= Solution
For a <dispersion relation> $\omega(k)$, <phase velocity> is $v_p=\omega/k$ and <group velocity> is $v_g=d\omega/dk=v_p+k\,dv_p/dk$. Their <local equality of phase and group velocity> holds throughout an interval for a nondispersive branch $\omega=ck$. Equality at one particular $k$ requires only that $v_p$ have zero <derivative> there, so it does not imply a globally nondispersive medium.
Use $k=2\pi/\lambda$ in the prescribed <gravity-capillary wave> relation. Then $\omega^2=gk+(T/\rho)k^3$, and differentiation gives
$$
\boxed{v_g=
\frac{g\lambda/(2\pi)+6\pi T/(\lambda\rho)}
{2\sqrt{g\lambda/(2\pi)+2\pi T/(\lambda\rho)}}.}
$$
In the gravity-dominated <limit>, $v_g=v_p/2$; in the surface-tension-dominated <limit>, $v_g=3v_p/2$. For $g,T,\rho>0$, the two are equal at $\lambda=2\pi\sqrt{T/(\rho g)}$, the minimum of the <phase velocity> for this dispersion relation.
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