I am doing a Runge-Kutta problem for unmarked homework, and have to do a second order taylor expansion of the function
$$f(x+\tfrac{2h}{3}, y + \tfrac{2h}{3}f(x+\tfrac{h}{3}, y + \tfrac{h}{3}f(x,y)))$$
for which I managed to get
$f(x,y) +\frac{2h}{3}\frac{\partial f}{\partial x} + \frac{2h}{3}\frac{\partial f}{\partial y} f(x,y) + ...\quad$ but the final terms seem to be elusive to me
I have some inkling that it's $\frac{h^2}{9}\frac{1}{2!}\frac{\partial^2f}{\partial x^2}+\frac{h^2}{9}\frac{1}{2!}\frac{\partial^2f}{\partial y^2}f^2(x,y)$ but when using these numbers for the rest of the question, I do not get the answer I require.
Thanks for the help
Note that the question says "Restrict yourself to the case of a single first-order ODE, that is, to the case m = 1"
Instead of a reduction to state-dimension one I would propose a reduction to an autonomous ODE. Then, all omitted arguments being the basis point, \begin{align} k_1&=f\\ k_2&=f(y+\tfrac h3k_1)\\ &=f+\tfrac{h}3f'f+\tfrac{h^2}{18}f''[f,f]+O(h^3)\\ k_3&=f(y+\tfrac{2h}3k_2)\\ &=f+\tfrac{2h}3f'k_2+\tfrac{2h^2}9f''[k_2,k_2]+O(h^3)\\ &=f+\tfrac{2h}3f'(f+\tfrac{h}3f'f)+\tfrac{2h^2}9f''[f,f]+O(h^3)\\ &=f+\tfrac{2h}3f'f+\tfrac{2h^2}9(f'f'f+f''[f,f])+O(h^3) \end{align} Where $f''[f,f]$ denotes the application of the vector-valued symmetric bi-linear form $f''$ to twice $f$ as arguments.
As you want to combine $$ y(t+h)=y+hf+\tfrac{h^2}2f'f+\tfrac{h^3}6(f'f'f+f''[f,f])+O(h^3) \\=y+h(b_1k_1+b_2k_2+b_3k_3)+O(h^3) $$ you get the conditions $b_1+b_2+b_3=1$, $\frac13b_2+\frac23b_3=\frac12$, $\frac29b_3=\frac16$ and $\frac1{18}b_2+\frac29b_3=\frac16$ resulting in $b_2=0$, $b_3=\frac34$, and $b_1=\frac14$.