= Solution
Assume $f$ is sufficiently smooth, for example $C^3$ near the exact trajectory. Write $f,f',f''$ at the current value, with $f'$ a <linear map> and $f''$ a bilinear map when the <ordinary differential equation> is vector valued. <Taylor expansion> of the stages gives
$$
\begin{aligned}
k_2&=f+ha_1f'f+\tfrac12h^2a_1^2f''[f,f]+O(h^3),\\
k_3&=f+h(a_2+a_3)f'f+h^2\left(a_1a_3(f')^2f+\tfrac12(a_2+a_3)^2f''[f,f]\right)+O(h^3).
\end{aligned}
$$
The exact flow has <Taylor expansion>
$$
y(t+h)=y+hf+\frac{h^2}{2}f'f+\frac{h^3}{6}\left(f''[f,f]+(f')^2f\right)+O(h^4).
$$
Matching the terms through $h^3$ in the <Runge-Kutta method> gives the sufficient <third-order conditions for an explicit Runge-Kutta method>
$$
\boxed{\begin{aligned}b_1+b_2+b_3&=1,\\b_2a_1+b_3(a_2+a_3)&=\tfrac12,\\b_2a_1^2+b_3(a_2+a_3)^2&=\tfrac13,\\b_3a_1a_3&=\tfrac16.\end{aligned}}
$$
For scalar $f$, replace $f''[f,f]$ by $f''f^2$ and $(f')^2f$ by $(f')^2f$ in the scalar sense. The resulting local error is $O(h^4)$; under a local <Lipschitz continuity> assumption and bounded derivatives on a finite time interval, the standard one-step error recurrence gives global error $O(h^3)$, hence third order.
On the linear test equation $y'=\lambda y$, putting $z=h\lambda$ gives the <stability function>
$$
R(z)=1+z+\frac{z^2}{2}+\frac{z^3}{6},\qquad R(-5/2)=-\frac{47}{48}.
$$
Thus $\boxed{|R(-5/2)|=47/48<1}$, so $-5/2$ lies inside the <linear stability domain>. The four order conditions are compatible, for example $a_1=1/2,a_2=-1,a_3=2$ and $(b_1,b_2,b_3)=(1/6,2/3,1/6)$.
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