= Solution
A comoving material cell contains the same particles in its perturbed and background descriptions, so its mass is $\bar\rho a^3d^3q=\rho d^3r$. The <Jacobian determinant> is $a^3\det(I+\nabla_q\psi)$. Expanding the <determinant> gives
$$
\frac\rho{\bar\rho}=\det(I+\nabla_q\psi)^{-1}
=1-\nabla_q\cdot\psi+O(\psi^2),\qquad
\boxed{\delta=-\nabla_q\cdot\psi}.
$$
Split the potential into homogeneous and perturbed parts. There is a sign error in the printed hint: $\nabla^2\Phi_0=4\pi G\bar\rho$ requires \b[$\nabla\Phi_0=+4\pi G\bar\rho r/3$], not the printed negative sign. Its acceleration $-\nabla\Phi_0$ agrees with $\ddot a/a=-4\pi G\bar\rho/3$.
For the perturbation, $\nabla_q^2\phi=4\pi G\bar\rho a^2\delta=-4\pi G\bar\rho a^2\nabla_q\cdot\psi$. Under the usual homogeneous boundary convention the curl-free force therefore satisfies
$$
\nabla_q\phi=-4\pi G\bar\rho a^2\psi_L,
$$
where $\psi_L$ is the longitudinal part of the displacement. Expanding the particle acceleration gives $\ddot r=\ddot a(q+\psi)+2\dot a\dot\psi+a\ddot\psi$. The background acceleration cancels the first term, leaving
$$
\ddot\psi+2H\dot\psi-4\pi G\bar\rho\psi_L=0,\qquad H=\dot a/a.
$$
For a longitudinal displacement this proves the displayed <longitudinal displacement in linear dust perturbations> equation with $\psi_L=\psi$. For unrestricted displacement, a transverse component has no <density contrast> and instead obeys $\ddot\psi_T+2H\dot\psi_T=0$; the full-vector equation printed in the question thus needs the longitudinal assumption. Taking the divergence removes this qualification and gives
$$
\boxed{\ddot\delta+2\frac{\dot a}{a}\dot\delta-4\pi G\bar\rho\delta=0}.
$$
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