= Solution
For the <Dahlquist test equation>, put $z=h\lambda$. The amplification roots satisfy
$$
(7-6z+2z^2)\xi^2-8\xi+1=0.
$$
We show that no root can reach the <unit circle> for $\operatorname{Re}z<0$. If $\xi=e^{i\theta}$, solving the quadratic in $z$ gives
$$
z=\frac32\pm\sqrt C,\qquad C=-\frac54+4e^{-i\theta}-\frac12e^{-2i\theta}.
$$
Write $c=\cos\theta$, $u=\operatorname{Re}C=-3/4+4c-c^2$, and $v=\operatorname{Im}C=(c-4)\sin\theta$. Then $u\leq9/4$ and
$$
\left(\frac92-u\right)^2-|C|^2
=\frac{81}{4}-9u-v^2
=(c-1)^2(c^2-6c+11)\geq0.
$$
Since $9/2-u>0$, this gives $|C|+u\leq9/2$. Consequently $|\operatorname{Re}\sqrt C|^2=(|C|+u)/2\leq9/4$, so both possible $z$ have nonnegative real part. Equality can occur only at $\theta=0$, which gives $z=0$ or $z=3$.
The leading coefficient has zeros $(3\pm i\sqrt5)/2$, both in the right half-plane. Hence the amplification roots vary continuously as a pair throughout the left half-plane. At $z=-1$ they are $1/3$ and $1/5$; neither can leave the <unit disk> without crossing the <unit circle>, which the preceding calculation excludes. The same conclusion extends to the imaginary axis, with strict inequality away from $z=0$. At zero the roots $1,1/7$ satisfy the simplicity requirement.
Thus the method is \b[<A-stable>]. Its third order does not contradict the usual second-order barrier, because it is a <multiderivative multistep method>.
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