= Solution
Differentiate the identity $f(u,g(u))=0$ using the <chain rule>:
$$
f_u+f_vg'(u)=0,\qquad f_u=-f_vg'(u).
$$
Substitution into the <trace> gives
$$
\operatorname{tr}J=-1+\frac{f_v}{\varepsilon}\bigl(1+\gamma g'(\mu)\bigr).
$$
Because $f_v<0$, the exact instability condition is
$$
\boxed{1+\gamma g'(\mu)<\frac\varepsilon{f_v},\qquad
g'(\mu)<-\frac1\gamma+\frac{\varepsilon}{\gamma f_v}.}
$$
For fixed $\delta>0$ and fixed <equilibrium> location, the correction tends to zero with $\varepsilon$. To leading order this is
$$
\boxed{g'(\mu)<-1/\gamma.}
$$
Thus instability requires the descending <nullcline> to be steeper than a line of slope $-1/\gamma$. The sign of the finite-$\varepsilon$ correction makes the exact unstable interval slightly smaller. Since $\delta=0.001$ can make $|f_v|$ small, the approximation quantitatively requires $\varepsilon/|f_v|\ll1$ at the point in question; merely comparing $\varepsilon$ with one can be insufficient.
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