 Solution

Differentiate the identity $f(u,g(u))=0$ using the chain rule:Substitution into the trace givesBecause $f_v<0$, the exact instability condition isFor fixed $\delta >0$ and fixed equilibrium location, the correction tends to zero with $\epsilon$. To leading order this isThus instability requires the descending nullcline to be steeper than a line of slope $-1/\gamma$. The sign of the finite-$\epsilon$ correction makes the exact unstable interval slightly smaller. Since $\delta =0.001$ can make $|f_v|$ small, the approximation quantitatively requires $\epsilon /|f_v|\ll 1$ at the point in question; merely comparing $\epsilon$ with one can be insufficient.