= Solution
Write $\boldsymbol\omega=\nabla\times\mathbf u$ and use the vector identity $(\mathbf u\cdot\nabla)\mathbf u=\nabla(|\mathbf u|^2/2)-\mathbf u\times\boldsymbol\omega$. Taking the <curl> of the momentum equation eliminates all pure <gradients>, including <gravity> and the $\nabla(\nabla\cdot\mathbf u)$ viscous contribution. Before invoking barotropy, the <pressure> term gives
$$
\nabla\times\left(-\frac1\rho\nabla p\right)=-\nabla(1/\rho)\times\nabla p=\frac{\nabla\rho\times\nabla p}{\rho^2}.
$$
This is <baroclinic vorticity generation>. For a <barotropic fluid>, $p=p(\rho)$, so $\nabla p$ is parallel to $\nabla\rho$ and this source vanishes. Constant $\nu$ permits the <curl> and <Laplacian> to commute. The result is
$$
\boxed{\frac{\partial\boldsymbol\omega}{\partial t}=\nabla\times(\mathbf u\times\boldsymbol\omega)+\nu\nabla^2\boldsymbol\omega}.
$$
No incompressibility assumption was used. Since $\nabla\cdot\boldsymbol\omega=0$, an equivalent form is
$$
\frac{D\boldsymbol\omega}{Dt}=(\boldsymbol\omega\cdot\nabla)\mathbf u-\boldsymbol\omega\,\nabla\cdot\mathbf u+\nu\nabla^2\boldsymbol\omega.
$$
The terms represent <vortex stretching>/tilting, compression/dilatation and viscous diffusion. In particular, removing the baroclinic source does not make the <vorticity> of a three-dimensional parcel constant. A spatially varying entropy-dependent $K$ in $p=K\rho^\gamma$ need not define a single barotropic law; the common-pressure-law assumption matters.
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