= Solution
Set $\mathbf k=(k_x,0,k_z)\ne0$. <Incompressibility> gives $(\mathbf k\cdot\tilde{\mathbf u})f'(\xi)e^{st}=0$. Since $f$ is nonconstant and differentiable, it follows that $\mathbf k\cdot\tilde{\mathbf u}=0$; therefore
$$
\boxed{\mathbf u'\cdot\nabla\mathbf u'=e^{2st}ff'(\tilde{\mathbf u}\cdot\mathbf k)\tilde{\mathbf u}=0.}
$$
The background has no <advection> of a $y$-independent perturbation. Its only remaining cross-advection term is $(\mathbf u'\cdot\nabla)\mathbf U=(-3\Omega u'_x/2+q\Omega u'_z)\mathbf e_y$. The velocity/time terms have spatial factor $f$, while $\nabla P'=\tilde P\mathbf k g'(\xi)e^{st}$. For nonzero <pressure> amplitude, $g'$ must therefore be a constant multiple of $f$; absorbing that multiple into $\tilde P$ gives $g'=f$. An additive spatial constant in $g$ is a <pressure> gauge. If $\tilde P=0$, there is no such restriction on the unused <pressure> profile.
Matching coefficients gives
$$
\boxed{\begin{aligned}
s\tilde u_x&=-k_x\tilde P/\rho_0+2\Omega\tilde u_y,\\
s\tilde u_y&=-\Omega\tilde u_x/2-q\Omega\tilde u_z,\\
s\tilde u_z&=-k_z\tilde P/\rho_0,\\
k_x\tilde u_x+k_z\tilde u_z&=0.
\end{aligned}}
$$
Because the quadratic perturbation term vanishes identically, these amplitudes give an <exact single-phase incompressible perturbation of a shear flow>, not merely a first-order truncation. Any sufficiently differentiable profile with $g'=f$ works locally; boundary conditions may restrict the profiles in a particular domain.
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