(11) Let $0<x<1,0<y<1$, and $(1-x y)^{2}=2(1-x)(1-y)$. Then the maximum value of the function $f(x, y)=\frac{1}{2} x y(1-x y)$ is $\qquad$ .
11 Let $t=x y$, then $t \in(0,1)$, so from
$$
(1-x y)^{2}=2(1-x)(1-y),
$$

we know
$$
(1-t)^{2}=2(1+t-(x+y))
$$
$$
\begin{array}{l}
\leqslant 2(1+t-2 \sqrt{x y}) \\
=2(1+t-2 \sqrt{t}) \\
=2(1-\sqrt{t})^{2},
\end{array}
$$

thus $1-t \leqslant \sqrt{2}(1-\sqrt{t})$, solving this yields $0 \leqslant t \leqslant 3-2 \sqrt{2}$. Therefore,
$$
f(x, y)=\frac{1}{2} x y(1-x y)=\frac{1}{2} t(1-t)
$$

has a maximum value of $5 \sqrt{2}-7$, at this point $x=y=\sqrt{2}-1$.