So I have to show that the single step method with the function $$ \Phi(t,y,h) = f(t,y) + \frac{h}{2}g(t + \frac{h}{3},y + \frac{h}{3} f(t,y)) ,$$ where $ g(t,y) = f_t(t,y) + f_y(t,y)f(t,y)$, has order 3.
So I know that :
I have to consider $\frac{1}{h}(y(t+h) - y(t)) - \Phi(t,y,h)$. Now I consider the taylor expansion of $y(t+h)$. \begin{align} y(t+h) &= y(t) + hy'(t) + \frac{h^2}{2} y''(t) + \frac{h^3}{6}y'''(t) + O(h^4)\\ &= y(t) +hf + \frac{h^2}{2}\frac{\delta f}{\delta t}f + \frac{h^3}{6} \frac{\delta^2 f}{\delta^2 t}f + O(h^4) \\ &= y(t) +hf + \frac{h^2}{2}(f_t + f_yf) + \frac{h^3}{6}( f_{tt} + f_{ty}f + f_{yy}f^2 + f_y(f_t + f_yf)) + O(h^4) \end{align} Now I can consider $$\frac{1}{h}(y(t+h) - y(t))- \Phi(t,y,h) = \frac{1}{h}(y(t+h) - y(t)) - f(t,y) + \frac{h}{2}g(t + \frac{h}{3},y + \frac{h}{3} f(t,y)) .$$ The expression " $y(t) + hf$ " of the Taylor expansion of $y(t+h)$ disappears, because of the factor $\frac{1}{h}$, the subtraction of $y(t)$ and $f$. So I only have to show that the Taylor expansion of $$\frac{h}{2}g(t + \frac{h}{3},y + \frac{h}{3} f(t,y)) $$ kill the expression " $ \frac{h^2}{2}(f_t + f_yf) + \frac{h^3}{6}( f_{tt} + f_{ty}f + f_{yy}f^2 + f_y(f_t + f_yf)) $ ", but I failed. Here I need your help.
Thank you for your time.
Let's reduce the consideration to autonomous systems $y'=f(y)$. Then expanding into a cubic Taylor polynomial and collapsing the last two terms as a linear Taylor expansion gives \begin{align} y(t+h)&=y(t)+hy'(t)+\frac{h^2}2 y''(t)+\frac{h^3}6y'''(t)+O(h^4) \\[.5em] &=y(t)+y'(t)h+\frac{h^2}2\left(y''(t)+\frac h3y'''(t)\right)+O(h^4) \\[.5em] &=y(t)+f(y(t))h+\frac{h^2}2 y''\left(t+\frac h3\right)+O(h^4) \end{align} Now using $y''(t)=g(y)=f'(y)f(y)$ one can return to the linear Taylor expansion in \begin{align} y''\left(t+\frac h3\right)&=g\left(y\Bigl(t+\frac h3\Bigr)\right) \\ &=g\left(y(t)+\frac h3y'(t)+O(h^2)\right) \\ &=g\left(y(t)+\frac h3f(y(t))\right)+O(h^2) \end{align} which is what the method computes $$ y(t+h)=y(t)+f(y(t))h+\frac{h^2}2 g\left(y(t)+\frac h3f(y(t))\right)+O(h^4) $$