= Solution
Use the <stability function> $R(z)=(z^2+6z+12)/(z^2-6z+12)$. Its denominator vanishes only at $3\pm i\sqrt3$, both in the open right half-plane. For $z=x+iy$,
$$
|z^2-6z+12|^2-|z^2+6z+12|^2
=-24x(|z|^2+12).
$$
This is nonnegative whenever $x\leq0$, so $|R(z)|\leq1$ throughout the closed left half-plane. The internal stage equations are also solvable there, since $\det(I-zA)=1-z/2+z^2/12$. Therefore \b[the method is <A-stable>].
The inequality is strict in the open left half-plane, while $|R(iy)|=1$. Moreover $R(z)\to1$ as $|z|\to\infty$, so the method is \b[not <L-stable>]. These conclusions follow directly from the definitions of <A-stability> and <L-stability>, rather than from the order of the method.
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