Is flipping a coin is a stable rotation? If I rotate a coin with dimensions: 10X10X1
about 1 of the big axes(10) in space, where there is no torque, will the rotation will be stable just like a frisbee or a football regular rotations are? please prove mathematically?
my proof tells me that this is not stable, am I wrong, where?
my proof: (assume coin is spheroid):

 A: For other readers, a relevant reference for the equations used above is the Wikipedia page on the 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.
A: 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!
