9. If the positive real numbers $x, y$ satisfy
$$
x^{3}+y^{3}=(4 x-5 y) y \text {, }
$$

then the maximum value of $y$ is . $\qquad$
9. $\frac{1}{3}$.
Let $x=k y$. Then $y=\frac{4 k-5}{k^{3}+1}$.
Let $f(k)=\frac{4 k-5}{k^{3}+1}$, then
$$
f^{\prime}(k)=-\frac{(k-2)\left(8 k^{2}+k+2\right)}{\left(k^{3}+1\right)^{2}} \text {. }
$$

Therefore, $f(k) \leqslant f(2)=\frac{1}{3}$.
Thus, the maximum value of $y$ is $\frac{1}{3}$.