根据题意有 $(2x-y)(x+y)=1,$ 考虑设$$\begin{cases} m=2x-y,\\n=x+y.\end{cases}$$则有 $mn=1$，且有$$\begin{cases} x=\dfrac13(m+n),\\y=\dfrac13(2n-m).\end{cases}$$记 $M=\dfrac{x-2y}{5x^2-2xy+2y^2}$，那么有$$\begin{split}M&=\dfrac{m-n}{m^2+n^2}\\&=\dfrac{m-n}{(m-n)^2+2}\\&=\dfrac{1}{m-n+\dfrac2{m-n}}\\&\leqslant\dfrac{\sqrt2}4.\end{split}$$当且仅当 $m-n=\sqrt2$ 即 $(x,y)=\left(\pm\dfrac{\sqrt6}3,-\dfrac{\sqrt2}2\pm\dfrac{\sqrt6}{6}\right)$ 时 $M$ 取得最大值 $\dfrac{\sqrt2}{4}$．