= Solution
\b[Yes: this is an A-stable third-order method.] Its second derivative is essential; the <Second Dahlquist barrier> applies to ordinary first-derivative <linear multistep methods>, not this <multiderivative multistep method>.
Apply the <Dahlquist test equation> $y'=\lambda y$ and put $z=h\lambda$. The amplification <polynomial> is
$$
Q(z)\zeta^2-8\zeta+1=0,\qquad Q(z)=7-6z+2z^2.
$$
We verify the <root condition for a multistep method> throughout $\operatorname{Re}z\le0$. For a quadratic $a\zeta^2+b\zeta+c$ with $|a|>|c|$, the <complex quadratic Schur criterion> is
$$
|\overline a b-c\overline b|<|a|^2-|c|^2
$$
for both roots to be strictly inside the <unit disk>. Here is a brief justification of the criterion rather than just a root plot. Set $P^\#(\zeta)=\zeta^2\overline{P(1/\overline\zeta)}$. The <polynomial> $\overline aP-cP^\#$ is $\zeta(D\zeta+E)$, where $D=|a|^2-|c|^2$ and $E=\overline a b-c\overline b$. Its two roots are inside precisely when $|E|<D$. On the <unit circle> $|P^\#|=|P|$, and $|c|<|a|$, so the <Rouche theorem> equates its interior root count with that of $P$. A boundary zero of $P$ would also be a boundary zero of the transformed <polynomial>, which the strict inequality excludes. This proves the strict criterion; the non-strict form follows by continuity, with any unit root simple because $|c/a|<1$.
For the present <polynomial> it is enough to show $8|Q-1|<|Q|^2-1$, except at $z=0$. Write $z=-r+iy$, $r\ge0$, and put $k=r^2+3r$, $s=y^2$. Then
$$
Q=(7+2k-2s)-i(6+4r)y,
\qquad X:=|Q|^2-1=(2k+6)(2k+8)+8(k+1)s+4s^2>0.
$$
Direct expansion gives the nonnegative factorization
$$
X^2-64|Q-1|^2
=16\left[k(k+3)^2(k+8)+4k(k^2+8k+11)s
+2(3k^2+11k+6)s^2+4(k+1)s^3+s^4\right].
$$
Every term is nonnegative, and the sum is strictly positive unless $r=y=0$. Also $|Q|^2=X+1>1$, so the criterion applies. For every nonzero $z$ in the closed left half-plane both amplification roots are strictly inside the <unit disk>. At $z=0$ they are $1$ and $1/7$, and the unit root is simple. Therefore
$$
\boxed{\text{the method is A-stable}.}
$$
This proof establishes stability of both the principal and parasitic modes, including on the imaginary axis.
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