= Solution
Along a smooth exact solution, the <chain rule> gives $g(y)=f'(y)f(y)=y''$. Expand the unscaled defect of the <multiderivative multistep method> about the central time $t_n$. The antisymmetric difference of the solution and second derivative, and the symmetric sum of first derivatives, give
$$
\begin{aligned}
E_h={}&(2-2\alpha-\gamma)h y'+\left(\frac13-\alpha+2\beta\right)h^3y^{(3)}\\
&+\left(\frac1{60}-\frac\alpha{12}+\frac\beta3\right)h^5y^{(5)}+\left(\frac1{2520}-\frac\alpha{360}+\frac\beta{60}\right)h^7y^{(7)}+O(h^9).
\end{aligned}
$$
All derivatives are evaluated at $t_n$. Reflection symmetry removes every even power. The fifth-order conditions require the coefficients of $h,h^3,h^5$ to vanish. The last two equations give $\beta=1/15$ and $\alpha=7/15$, and the first then gives $\gamma=16/15$. Thus
$$
\boxed{\alpha=\frac7{15},\qquad\beta=\frac1{15},\qquad\gamma=\frac{16}{15}.}
$$
These determine the <symmetric two-step two-derivative formula>. Substitution into the next coefficient gives
$$
E_h=\frac{h^7}{4725}y^{(7)}+O(h^9).
$$
\b[The requested fifth-order conditions actually produce formal order six.] In the usual maximal-order terminology there is no member of this symmetric family with exact order five: cancelling the fifth-power defect also cancels the automatically zero sixth-power defect. Interpreting the printed request as order at least five gives precisely the boxed parameters.
At zero step size the recurrence polynomial is $w^2-1$, with simple roots $1$ and $-1$, so it satisfies the <root condition for a multistep method> and is <zero-stable>. With sufficiently smooth locally Lipschitz derivative maps, the nearby implicit solution branch and starting values of matching accuracy, the <convergence of a zero-stable multiderivative method> gives global order six. Formal order by itself does not imply decay of its additional numerical mode.
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