= Solution
Along the exact <ordinary differential equation> solution, the <chain rule> gives $g(y)=f'(y)f(y)=y''$. Insert the exact solution into the <symmetric two-step two-derivative formula> and expand about the middle time $t=t_{n+1}$. Its unnormalized <local truncation error> is
$$
\begin{aligned}
\delta_h={}&y(t+h)-y(t-h)-\frac{7h}{15}[y'(t+h)+y'(t-h)]-\frac{16h}{15}y'(t)\\
&+\frac{h^2}{15}[y''(t+h)-y''(t-h)].
\end{aligned}
$$
The coefficients of $hy'$, $h^3y'''$ and $h^5y^{(5)}$ are respectively
$$
2-\frac{14}{15}-\frac{16}{15}=0,\qquad\frac13-\frac7{15}+\frac2{15}=0,\qquad\frac1{60}-\frac7{180}+\frac1{45}=0.
$$
The centered residual is an <odd function> of $h$, so every even power cancels. The next coefficient is
$$
\frac1{2520}-\frac7{5400}+\frac1{900}=\frac1{4725},\qquad\delta_h=\frac{h^7}{4725}y^{(7)}(t)+O(h^9).
$$
Thus \b[the method has order six]. Its <zero-stability> roots at $h=0$ are the simple roots $1,-1$. With sufficiently accurate starts and smooth derivative evaluation maps, <convergence of a zero-stable multiderivative method> gives sixth-order error on fixed time intervals. The <first Dahlquist barrier> for ordinary <linear multistep methods> does not apply, because this is a <multiderivative multistep method>.
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