= Solution
Retain $a=p_0^{1/\gamma}/\rho_0$ from part (b), and use time translation invariance to write $G(\mathbf x,t;\mathbf y,\tau)=g(\mathbf x,\mathbf y;t-\tau)$. Define the temporal <Fourier transform> by
$$
\widetilde G(\mathbf x,\mathbf y;\omega)=\int_{\mathbb R}g(\mathbf x,\mathbf y;s)e^{-i\omega s}\,ds.
$$
Multiplying the transformed <Green function> equation by $-a(\mathbf x)$ gives
$$
\nabla\cdot(a\nabla\widetilde G)+\frac{a\omega^2}{c_0^2}\widetilde G
=-a(\mathbf y)\delta(\mathbf x-\mathbf y).
$$
This is a symmetric divergence-form spatial operator, although the original unweighted operator need not be symmetric in the ordinary volume measure.
Let $G_j(\mathbf x)=\widetilde G(\mathbf x,\mathbf y_j;\omega)$. The product rule gives the bilinear <Green second identity>
$$
\nabla\cdot\{a(G_1\nabla G_2-G_2\nabla G_1)\}
=G_1\nabla\cdot(a\nabla G_2)-G_2\nabla\cdot(a\nabla G_1).
$$
The frequency terms cancel. Integrating over the domain, the right side becomes
$$
-a(\mathbf y_2)\widetilde G(\mathbf y_2,\mathbf y_1;\omega)
+a(\mathbf y_1)\widetilde G(\mathbf y_1,\mathbf y_2;\omega).
$$
The boundary integral is zero for common homogeneous <Dirichlet boundary condition>, <Neumann boundary condition> or reciprocal <Robin boundary condition> conditions. In an unbounded domain, use the same outgoing limiting-absorption prescription for both <Green functions>; it gives the corresponding vanishing boundary pairing. This identity has no complex conjugation: it proves transpose <wave reciprocity>, not a Hermitian or time-reversal identity. We obtain the <weighted acoustic Green-function reciprocity>
$$
\widetilde G(\mathbf y_1,\mathbf y_2;\omega)
=\frac{a(\mathbf y_2)}{a(\mathbf y_1)}\widetilde G(\mathbf y_2,\mathbf y_1;\omega).
$$
The factor is frequency independent, so inverse <Fourier transform> gives the same relation at equal time lag. Both time arguments below have lag $t-\tau_2$, hence
$$
\boxed{G(\mathbf y_1,t;\mathbf y_2,\tau_2)
=G(\mathbf y_2,t+\tau_1-\tau_2;\mathbf y_1,\tau_1)
\frac{[p_0(\mathbf y_2)]^{1/\gamma}\rho_0(\mathbf y_1)}{[p_0(\mathbf y_1)]^{1/\gamma}\rho_0(\mathbf y_2)}.}
$$
Reciprocity exchanges source and receiver while preserving elapsed time; it does not turn a causal response into an advanced one.
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