7. If positive real numbers $x, y$ satisfy $x(x+2 y)=9$, then the maximum value of $x^{5} y$ is $\qquad$

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$x^{2}+2 x y=9 \Rightarrow y=\frac{9-x^{2}}{2 x}$, then $x^{5} y=\frac{1}{2}\left(9 x^{4}-x^{6}\right)$. Let $f(x)=\frac{1}{2}\left(9 x^{4}-x^{6}\right)$, $f^{\prime}(x)=3 x^{3}\left(6-x^{2}\right)$, so when $x=\sqrt{6}$, $x^{5} y$ reaches its maximum value of 54.