Let $t = xy$, then
$$x^2 + y^2 = 1 - 2t,$$
$$x^3 + y^3 = 1 - 3t,$$
$$x^4 + y^4 = 1 - 4t + t^2$$
$$x^5 + y^5 = 1 - 5t + 5t^2.$$
Since $x^2 + y = x + y^2$, the original inequality is equivalent to
$$\begin{aligned} & \frac{x}{x^2 + y^3} + \frac{y}{x^3 + y^2} \le \frac{2}{x+y^2} \\ \Leftrightarrow & \frac{x^4 + y^4 + xy}{(x^2 + y^3)(x^3 + y^2)} \le \frac{4}{x+y+x^2+y^2} \\ \Leftrightarrow & (1+x^2+y^2)(x^4+y^4+xy) \le 4(x^2+y^3)(x^3+y^2) \\ \Leftrightarrow & (2-2t)(1-3t+2t^2) \le 4(1-5t+6t^2+t^3) \\ \Leftrightarrow & (4t-1)(t^2+2t-1) \ge 0 \\ \Leftrightarrow & (t-\frac{1}{4})(t-(\sqrt{2}-1))(t+\sqrt{2}+1) \ge 0. \end{aligned}$$
From
$$0 < t \le \left(\frac{x+y}{2}\right)^2 = \frac{1}{4}$$
we know that the last inequality above holds.