= Solution
Vary the action with compactly supported variations and integrate by parts. Its overall normalization does not affect the equation of motion:
$$
\partial_\mu(\sqrt{-g}\,g^{\mu\nu}\partial_\nu\phi)=0.
$$
Here $\sqrt{-g}=e^{3t}$, so the <massless scalar field> obeys
$$
\phi_{tt}+3\phi_t-e^{-2t}\nabla^2\phi=0.
$$
Separate a spatial <Fourier mode> as $\phi(t,\mathbf x)=F_k(t)e^{i\mathbf k\cdot\mathbf x}$. Its mode equation is $F_{k,tt}+3F_{k,t}+k^2e^{-2t}F_k=0$. With <conformal time> $\tau=-e^{-t}$, $a=-1/\tau$ and $\mathcal H=-1/\tau$, it becomes
$$
F_k''-\frac2\tau F_k'+k^2F_k=0.
$$
For $k>0$, put $z=-k\tau=ke^{-t}$ and $F_k=z^2G(z)$. Direct differentiation reduces the equation to
$$
G_{zz}+\frac2zG_z+\left(1-\frac2{z^2}\right)G=0,
$$
the <Spherical Bessel function> equation with order one. Hence the complete classical mode basis is
$$
\boxed{F_k(t)=z^2[C_{\mathbf k}j_1(z)+D_{\mathbf k}y_1(z)],\qquad z=ke^{-t}.}
$$
Since $z^2j_1(z)=\sin z-z\cos z$ and $z^2y_1(z)=-\cos z-z\sin z$, an equivalent complex basis is
$$
\boxed{F_k(\tau)=A_{\mathbf k}(1+ik\tau)e^{-ik\tau}
+B_{\mathbf k}(1-ik\tau)e^{ik\tau}.}
$$
The classical action does not select one frequency branch or a quantum vacuum. A real field has conjugate-related coefficients at opposite wavevectors. The zero mode, which is not obtained by treating these two $k>0$ basis functions as independent at $k=0$, is
$$
\boxed{F_0(t)=C_0+D_0e^{-3t}.}
$$
At late times one independent solution freezes and the other decays as $e^{-3t}$. These are minimally coupled massless modes; the supplied action has no curvature-coupling term, so it should not be replaced by the conformally coupled wave equation.
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