 Solution

A nonzero spatially constant steady state satisfies $v=u$ and $v=u^2$, hence $(u,v)=(1,1)$. The first derivatives of $u^2/v-u$ there are $1,-1$, while those of $Q(u^2-v)$ are $2Q,-Q$. ThusSince $Q>0$, both eigenvalues have strictly negative real part exactly when $Q>1$. For $0<Q<1$ the equilibrium is unstable. At $Q=1$ its linear eigenvalues are $\pm i$; this is the marginal boundary and does not give asymptotic stability from linearization. The positive-trace criterion in the hint tests positive real parts; stability requires applying it to $-J$.