(5) Given that $x, y$ are both positive real numbers, then the maximum value of $\frac{x}{2 x+y}+\frac{y}{x+2 y}$ is ( ).
(A) 2
(B) $\frac{2}{3}$
(C) 4
(D) $\frac{4}{3}$
$5 \mathrm{~B}$ Hint: Solution 1 Let $s=2 x+y, t=x+2 y$, then
$$
x=\frac{1}{3}(2 s-t), y=\frac{1}{3}(2 t-s) .
$$

Therefore,
$$
\frac{x}{2 x+y}+\frac{y}{x+2 y}=\frac{4}{3}-\frac{1}{3}\left(\frac{t}{s}+\frac{s}{t}\right) \leqslant \frac{2}{3} .
$$

Solution 2 Let $t=\frac{y}{x}$, then $t \in(0,+\infty)$, at this time
$$
\frac{x}{2 x+y}+\frac{y}{x+2 y}=\frac{1}{t+2}+\frac{t}{2 t+1}=f(t),
$$

which gives
$$
f^{\prime}(t)=-\frac{3\left(t^{2}-1\right)}{(t+2)^{2}(2 t+1)^{2}} .
$$

Clearly, when $t<1$, $f^{\prime}(t)>0$; when $t>1$, $f^{\prime}(t)<0$, so the function $f(t)$ reaches its maximum value $f(1)=\frac{2}{3}$ at $t=1$, i.e., $x=y$.