= Solution
Eliminate $\dot v$ from the first equation before <linearization>:
$$
\dot u=\mu-u-\frac\gamma\varepsilon f(u,v),\qquad
\dot v=\frac1\varepsilon f(u,v).
$$
All partial <derivatives> below are evaluated at $(\mu,g(\mu))$. The <Jacobian matrix> is
$$
J=\begin{pmatrix}
-1-\gamma f_u/\varepsilon&-\gamma f_v/\varepsilon\\
f_u/\varepsilon&f_v/\varepsilon
\end{pmatrix}.
$$
Its <trace> and <determinant> are
$$
\operatorname{tr}J=-1+\frac{f_v-\gamma f_u}{\varepsilon},\qquad
\det J=-\frac{f_v}{\varepsilon}>0.
$$
By <linear stability of a planar equilibrium>, a positive <trace> makes both real <eigenvalues> positive or gives a complex pair with positive real part. The <equilibrium> is consequently unstable precisely when
$$
\boxed{\varepsilon-f_v<-\gamma f_u.}
$$
A negative <trace> gives <asymptotic stability>; equality leaves a purely imaginary pair and requires a separate nonlinear <bifurcation> calculation. In particular, the unique positive <equilibrium> is never a <saddle equilibrium>.
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