Question: I'm reading a book in which it says that a system is at mechanical equilibrium if the pressure is *constant in time but not necessarily uniform in space* at any fixed point (for example it can be caused by the gravitational field). Given this definition of mechanical equilibrium, it is clear that you can't define an equilibrium pressure with a scalar; in general it is a stationary field: $p=p(\mathbf{x})$. But when the book draws a quasi-static process, it draws it with a well defined line, so implicitly it is considering pressure as a scalar at the equilibrium. This is a first point I'd like to clarify.

Second, it is also written that in a state of non equilibrium, the *pressure is not defined* for a system. I don't understand why, since fixed the time and a point it must be defined the pressure there. Of course the value of the pressure in this point will change in time but I can't get why you can never define it.

Answer: 
For other readers, a relevant reference for the equations used above is the Wikipedia page on the [tennis racket theorem](https://en.wikipedia.org/wiki/Tennis_racket_theorem).

Your proof demonstrates that, to first order, the stability is marginal. This is what you showed by demonstrating that (to first order) $d \omega\_x' = d \omega\_z' = 0$. That is, it is on the border of stable and unstable. Any initial perturbation will neither grow nor shrink but will stay the same size.

At the end you show that $d \omega\_y'$ is given by a constant that has the same sign as $d \omega\_z$. I could not find an error in the derivation after a few minutes, so although I am not completely sure it is correct, it probably is. The conclusion is probably still incorrect, though. A perturbation that grows linearly in time might still be called marginally stable by most physicists. It is still quite different than the usual stable/unstable cases, where the perturbation will either grow or shrink exponentially in time.
with the "tennis racket theorem" you get this differential equation:

$\ddot{\omega}\_y=-k\,\omega\_y$

so the solution is: (with the initial condition $\omega\_y(0)=A\,,\quad D(\omega)\_y(0)=0$ )

$\omega\_y(t)=A\cos(\sqrt{k}) $

if $k \ge 0$ you get a stable solution.

if $k < 0$ the solution is unstable ( $\omega\_y(t)=A\,\cosh(\sqrt{k}\,t)$

with:

$k={\frac {{\Omega\_{{x}}}^{2} \left( -I{\it yy}+{\it Ixx} \right) 
 \left( {\it Ixx}-{\it Izz} \right) }{{\it Iyy}\,{\it Izz}}}
$

your case:

$Ixx=Iyy=\frac{M\*(b^2+s^2)}{5}$

$Izz=\frac{2\,M\,b^2}{5}$

$\rightarrow$

$k=0$ stable system!