\[ \begin{aligned} \eta=\hat{\eta}\,e^{i(kx+ly-\omega t)} &\mathrel{\underset{\text{compute derivatives}}{\Rightarrow}}\; \frac{\partial\eta}{\partial t}=-i\omega\eta,\quad \frac{\partial\eta}{\partial x}=ik\eta,\quad \frac{\partial^2\eta}{\partial x^2}=-k^2\eta,\quad \frac{\partial^2\eta}{\partial y^2}=-l^2\eta \\ &\mathrel{\underset{\text{substitute}}{\Rightarrow}}\; \frac{\partial}{\partial t}\!\left[(-k^2-l^2-\tfrac{f_0^2}{c^2})\eta\right]+\beta(ik\eta)=0 \\ &\mathrel{\underset{\text{apply }\frac{\partial}{\partial t}}{\Rightarrow}}\; (-k^2-l^2-\tfrac{f_0^2}{c^2})(-i\omega\eta)+i\beta k\,\eta=0 \\ &\mathrel{\underset{\text{factor }i\eta\neq0}{\Rightarrow}}\; i\eta\Big[\omega\!\left(k^2+l^2+\tfrac{f_0^2}{c^2}\right)+\beta k\Big]=0 \\ &\mathrel{\underset{\text{set bracket }=0}{\Rightarrow}}\; \omega\!\left(k^2+l^2+\tfrac{f_0^2}{c^2}\right)+\beta k=0 \\ &\mathrel{\underset{\text{solve for }\omega}{\Rightarrow}}\; \boxed{\displaystyle \omega=-\frac{\beta k}{\,k^2+l^2+\tfrac{f_0^2}{c^2}\,}} \end{aligned} \]

(With $\omega>0$ by convention, the signs of $k$ and $l$ set the phase-propagation direction)

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