= Solution
Applying the stage equations to $y'=\lambda y$ gives the <stability function>
$$
R(z)=1+zb^T(I-zA)^{-1}\mathbf1
=\frac{1+z/2+z^2/12}{1-z/2+z^2/12},\qquad z=h\lambda.
$$
The poles are $3\pm i\sqrt3$, both in the right half-plane. To determine the whole domain rather than only its boundary, multiply numerator and denominator by 12 and put $z=x+iy$. Direct expansion gives
$$
|z^2-6z+12|^2-|z^2+6z+12|^2=-24x(|z|^2+12).
$$
Thus $|R(z)|\leq1$ exactly when $x\leq0$, with no pole in that set. Consequently
$$
\boxed{\mathcal D=\{z:\operatorname{Re}z\leq0\},\qquad\text{the method is A-stable}.}
$$
On the imaginary axis $|R|=1$, and in the open left half-plane it is less than one. Its limit along the negative real axis is one, so <A-stability> here does not imply stiff decay or <L-stability>.
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