= 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 $2Q,-Q$. Thus
$$
\boxed{J=\begin{pmatrix}1&-1\\2Q&-Q\end{pmatrix},\qquad
\operatorname{tr}J=1-Q,\quad\det J=Q.}
$$
Since $Q>0$, both <eigenvalues> have strictly negative real part exactly when \b[$Q>1$]. 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 $-J$.
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