= Solution
For any <vector> $\mathbf a$, differentiating its components and the rotating basis gives the <rotating-frame derivative formula>
$$
\left(\frac{d\mathbf a}{dt}\right)_S=\left(\frac{d\mathbf a}{dt}\right)_{S'}+\boldsymbol\omega\times\mathbf a.
$$
Apply it twice to the <position> $\mathbf x$, with constant <angular velocity> $\boldsymbol\omega$. The inertial <acceleration> is
$$
\mathbf a_S=\ddot{\mathbf x}+2\boldsymbol\omega\times\dot{\mathbf x}+\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf x),
$$
where dots denote rotating-frame derivatives. <Newton's second law> therefore becomes
$$
\boxed{m\ddot{\mathbf x}=\mathbf F-2m\boldsymbol\omega\times\dot{\mathbf x}-m\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf x).}
$$
The last terms are the <Coriolis force> and the force corresponding to <centrifugal acceleration>.
For the <particle on a uniformly rotating inclined plane>, choose orthonormal vectors $\mathbf e_1$ along the horizontal $\xi$ axis and $\mathbf e_2$ up the slope, and set $\mathbf e_3=\mathbf e_1\times\mathbf e_2$, the upward <normal vector>. Then $\mathbf e_z=\sin\theta\,\mathbf e_2+\cos\theta\,\mathbf e_3$ and
$$
\mathbf x=\xi\mathbf e_1+\eta\mathbf e_2,\qquad\boldsymbol\omega=\omega\mathbf e_z,\qquad\mathbf F=-mg\mathbf e_z+N\mathbf e_3.
$$
Here $N$ is the <normal reaction>. The marble is modeled as the smooth sliding particle specified by the printed equations, without an additional rolling constraint. The rotating-frame <Coriolis acceleration> and <centrifugal acceleration> are respectively
$$
-2\boldsymbol\omega\times\dot{\mathbf x}=2\omega\dot\eta\cos\theta\,\mathbf e_1-2\omega\dot\xi\cos\theta\,\mathbf e_2+2\omega\dot\xi\sin\theta\,\mathbf e_3,
$$
$$
-\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf x)=\omega^2\xi\mathbf e_1+\omega^2\eta\cos^2\theta\,\mathbf e_2-\omega^2\eta\sin\theta\cos\theta\,\mathbf e_3.
$$
The normal rotating-frame <acceleration> is zero, so the <normal reaction> is not generally $mg\cos\theta$; it satisfies
$$
N=m\bigl(g\cos\theta-2\omega\dot\xi\sin\theta+\omega^2\eta\sin\theta\cos\theta\bigr).
$$
Projecting the same equation onto the two tangential axes gives exactly
$$
\boxed{\ddot\xi=\omega^2\xi+2\omega\dot\eta\cos\theta,\qquad\ddot\eta=\omega^2\eta\cos^2\theta-2\omega\dot\xi\cos\theta-g\sin\theta.}
$$
Multiply the equations by $\dot\xi$ and $\dot\eta$ and add. The <Coriolis force> terms cancel, as expected because this force does no work relative to the rotating frame. The <normal reaction> also does no work along the plane. The conserved quantity is \b[rotating-frame <kinetic energy> plus gravitational and <centrifugal potential> energy]:
$$
\boxed{E'=\frac m2(\dot\xi^2+\dot\eta^2)+mg\eta\sin\theta-\frac{m\omega^2}{2}(\xi^2+\eta^2\cos^2\theta)=\text{constant}.}
$$
This need not be the conserved inertial <energy>, since the externally driven rotating plane can exchange energy with the particle.
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