Suppose that a method is given by $$y_{n+2/3} = y_n + h((1/3) f(y_{n+2/3}) + (1/3) f(y_{n}))$$ and $$y_{n+1} = y_n + h((3/4) f(y_{n+2/3}) + (1/4) f(y_{n})).$$ I am trying to obtain the Butcher tableau for this method. The main issue that I am having is how to deal with the $y_{n+2/3}$.
Are there any general guidelines for how to approach this? I considered using the substitution $y_{n+2/3} \approx (1/3) y_{n} + (2/3) y_{n+1}$, but this only seemed to make the problem more complex.
Define $$ k_1=f(y_n), ~~~ k_2=f(y_{n+2/3}) $$ and rewrite your equations as $$ y_{n+2/3}=y_n+\frac{h}3(k_1+k_2)\implies k_2=f\left(y_n+\frac{h}3(k_1+k_2)\right) $$ to eliminate $y_{n+2/3}$ completely. This should now look familiar enough.