This is one of the exercise questions in the book Numerical Analysis by Richard L.Burden
Show that the difference method $$y_0 = \alpha \\ y_{i+1} = y_i + a_1 f(t_i,y_i)+a_2 f(t_i+\alpha_2, y_i+\delta_2 f(t_i,y_i)$$ for each $i=0,1,2,\cdots,N-1$, cannot have local truncation error $\mathcal{O}(h^3)$ for any choice of constants $a_1,a_2,\alpha_2,$ and $\delta_2$
Answer:
Using the notation $y_{i+1}=y(t_{i+1}),y_i = y(t_i), $ and $f_i=f(t_i,y(t_i))$ , and also the definition of the local truncation error we have $$h\tau_{i+1}=y_{i+1}-y_i-a_1f_i-a_2f(t_i+\alpha_2,y_i+\delta_2 f_i))$$ If we expand $y_{i+1}$ and $f(t_i+\alpha_2,y_i+\delta_2 f_i)$ in Taylor series about $t_i$ and $f(t_i,y_i)$: $$\begin{equation}\begin{aligned} h\tau_{i+1} &= \left( y_i+hf_i + \frac{h^2}{2}f'_i+\frac{h^3}{6}f''_i+\cdots \right) - y_i - a_1f_i \\&- a_2 \left( f_i+\alpha_2 f_t(t_i,y_i)+\delta_2 f_i f_y (t_i,y_i) + \alpha_2 \delta_2 f_i f_{ty}(t_i,y_i) +\frac{\alpha^2}{2}f_{tt}(t_i,y_i) \\+ \frac{\delta^2 f_i^2}{2}f_{yy}(t_i,y_i)+\cdots\right) \\&= (h-a_1-a_2)f_i+ \left( \frac{h^2}{2}-a_2 \alpha_2 \right)f_t (t_i,y_i) + \left( \frac{h^2}{2}-a_2 \delta_2 \right)f_i f_y (t_i,y_i) \\&+\left( \frac{h^3}{6}-a_2\frac{\alpha^2}{2} \right)f_{tt}(t_i,y_i)+\left( \frac{h^3}{3}-a_2 \alpha_2 \delta_2 \right)f_i f_{ty} (t_i, y_i) \\&+ \left( \frac{h^3}{6} -a_2 \frac{\delta_2^2}{2}\right)f_i^2 f_{yy} (t_i,y_i)+\color{red}{\frac{h^3}{6}\left( f_t (t_i,y_i) f_y (t_i,y_i) + f_i f_y^2 (t_i,y_i) \right)}+\cdots \end{aligned}\end{equation}$$
I can only expand the equation until the black part, I couldn't get where the red part comes from, and it's the essence to show that the $h^3$ term cannot be cancelled regardless of the choice of the constants $a_1,a_2,\alpha_2,\delta_2$. The $\frac{h^3}{6}$ in the $\color{red}{\frac{h^3}{6}\left( f_t (t_i,y_i) f_y (t_i,y_i) + f_i f_y^2 (t_i,y_i) \right)}$ hinting that it comes from the Taylor expansion of $\frac{h^3}{6}f_i''$, but when I expand it I only get three terms: $$\frac{h^3}{6}f''_i = \frac{h^3}{6} \left( f_{tt}+f_{yy}f_i^2+2f_{ty}f_i \right)$$
$f''$ denotes the second derivative of $f(t, y(t))$ as function in $t$. With $y'(t)=f(t,y(t))$ it computes as $$ \frac{d^2}{dt^2}f\bigl(t,y(t)\bigr) =\frac{d}{dt}\Bigl(f_t\bigl(t,y(t)\bigr)+f_y\bigl(t,y(t)\bigr)f\bigl(t,y(t)\bigr)\Bigr) =f_{tt}+2f_{ty}f+f_{yy}f^2+f_y(f_t+f_yf). $$ So indeed there are two additional parts more than you expected. They result conceptually from the fact that the offset $y(t+s)-y(t)$ for the Taylor expansion at time $t$ is not constant but itself a function that contributes higher order terms. Directly they result from the application of the product rule to the second term in the first derivative.