= Solution
Use the stated <Fourier series> normalization, so $g(\boldsymbol\theta,\mathbf J,t)=\sum_{\mathbf n}g_{\mathbf n}(\mathbf J,t)e^{i\mathbf n\cdot\boldsymbol\theta}$. Then <3/b> gives
$$
\partial_t\delta f_{\mathbf n}+i\mathbf n\cdot\boldsymbol\Omega\,\delta f_{\mathbf n}-i\mathbf n\cdot\frac{\partial f_0}{\partial\mathbf J}\,\delta\Phi_{\mathbf n}^{\mathrm{tot}}=0.
$$
For the <Laplace transform> with kernel $e^{i\omega t}$, <integration by parts> yields
$$
\int_0^\infty e^{i\omega t}\partial_t\delta f_{\mathbf n}\,dt=-\delta f_{\mathbf n}(0)-i\omega\widetilde{\delta f}_{\mathbf n}.
$$
The upper boundary vanishes for sufficiently large $\operatorname{Im}\omega>0$, and the initial term is zero by hypothesis. Thus
$$
i(\mathbf n\cdot\boldsymbol\Omega-\omega)\widetilde{\delta f}_{\mathbf n}=i\mathbf n\cdot\frac{\partial f_0}{\partial\mathbf J}\,\widetilde{\delta\Phi}_{\mathbf n}^{\mathrm{tot}},
$$
or
$$
\boxed{\widetilde{\delta f}_{\mathbf n}(\mathbf J,\omega)=-\mathbf n\cdot\frac{\partial f_0}{\partial\mathbf J}\frac{\widetilde{\delta\Phi}_{\mathbf n}^{\mathrm{tot}}(\mathbf J,\omega)}{\omega-\mathbf n\cdot\boldsymbol\Omega(\mathbf J)},\qquad \boldsymbol\Omega=\frac{\partial H_0}{\partial\mathbf J}.}
$$
The denominator identifies the <orbital resonance> between the perturbation and unperturbed orbital <frequencies>. The zero harmonic has no driving term and remains zero with these initial data.
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