We consider this scalar IVP $\dot{y}(t)=f(t, y(t)), y(0)=y_0$ with Butcher Tableau
\begin{array} {c|cccc} 0 & 0 & 0 & 0\\ \frac{1}{3} & \frac{1}{3} & 0 & 0\\ \frac{2}{3} &\frac{2}{3} & 0 & 0 \\ \hline & 0 &\frac{1}{2} &\frac{1}{2} \end{array}
We're asked to find out the order of consistency $p$.
We first prove that this system is autonomization invariant. To do so, we need to check whether the sum of the elements of the right side of the three "upper rows" is equal to the element on the left. That is, whether $0 = 0 + 0 + 0$, $\frac{1}{3} = \frac{1}{3} + 0 + 0$, $\frac{2}{3} = \frac{2}{3} + 0 + 0$, which is the case. If a system is autonomization invariant, we do not need to consider the partial derivatives with respect to time while finding the derivatives we need. That is similar to Find $\alpha$ such that $y_{j+1}=y_j+\frac{h}{2 \alpha}k_1 + h(1- \frac{1}{2 \alpha})k_2$ has order of consistency 2 . There, it was assumed that the system is autonomization invariant, something we should have proved first similar to here.
Now, we need to expand $$\tau_{j+1} = (\dot{y}(t_j) - \Phi(t_j, y(t_j), 0)) $$$$+ \frac{h}{2}(\ddot{y}(t_j)- 2 \dot{\Phi}(t_j, y(t_j), 0)) + \frac{h^2}{6}(\dddot{y}(t_j)- 3 \ddot{\Phi}(t_j, y(t_j), 0)) $$$$+ ... + O(h^p)$$ until the first term in the parenthesis that is non-zero.
Let $$\dot{y}(t) = f(t, y(t))$$ then $$\ddot{y}(t) = \frac{df}{dy}(y(t))f(y(t))$$ and $$\dddot{y}(t) = \frac{d^2f}{dy^2}(y(t))f(y(t))^2 + (\frac{df}{dy}(y(t)))^2f(y(t))$$.
Let $$\Phi(h) = f(y(t) + \frac{1}{3}hf + \frac{2}{3}hf) $$
$$+ h\frac{1}{2}(\frac{df}{dy}(y(t))(\frac{1}{3} + \frac{2}{3})f(y(t) + \frac{1}{3}hf + \frac{2}{3}hf)) $$
$$+ \frac{h^2}{2} \frac{1}{2}(\frac{d^2f}{dy^2}(y(t)) (\frac{1}{3} + \frac{2}{3})^2 f(y(t) + \frac{1}{3}hf + \frac{2}{3}hf)^2) + ...$$ where
$$\Phi(y(t),0) = f(y(t))$$, $$\dot{\Phi}(y(t),0)= \frac{1}{2}(\frac{df}{dy}(y(t))f(y(t)))$$ and $$\ddot{\Phi}(y(t),0)= \frac{h^2}{2} \frac{1}{2}(\frac{d^2f}{dy^2}(y(t)) f(y(t))^2)$$
From that, we can see that $\Phi(h)= \dot{y}(t)$ and $\dot{\Phi}(y(t),0)= \ddot{y}(t)$ but $\ddot{\Phi}(y(t),0) \neq \dddot{y}(t)$. We can thus conclude that the order of consistency $p=2$, and not higher.