= Solution
Apply the method to the <Dahlquist test equation> $y'=\lambda y$, so $g(y)=\lambda^2y$, and put $z=h\lambda$. A trial numerical mode $y_n=w^n$ satisfies
$$
F_z(w)=(z^2-7z+15)w^2-16zw-(z^2+7z+15)=0.
$$
For every real $z<0$, the leading coefficient is positive, while
$$
F_z(-1)=2z<0,\qquad\lim_{w\to-\infty}F_z(w)=+\infty.
$$
The <intermediate value theorem> gives a real amplification root $w<-1$. Its modulus exceeds one although the exact scalar solution is decaying. Therefore
$$
\boxed{\text{The method is not A-stable; it is unstable for every negative real }z.}
$$
For a concrete choice, $z=-1$ gives $23w^2+16w-9=0$, whose negative root is $(-8-\sqrt{271})/23<-1$. This proves <negative-real instability of the symmetric two-derivative formula> directly, without using either an order barrier or the supplied quadratic criterion. The unstable branch is the <parasitic amplification root> issuing from $w=-1$ at $z=0$; its small-step expansion is $w=-1+z/15+O(z^2)$, so a negative $z$ moves it outside the unit circle immediately.
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