II. (40 points) Given positive real numbers $x, y, z$ satisfying $x+y+z=1$. Try to find the minimum value of the real number $k$ such that the inequality
$$
\frac{x^{2} y^{2}}{1-z}+\frac{y^{2} z^{2}}{1-x}+\frac{z^{2} x^{2}}{1-y} \leqslant k-3 x y z
$$

always holds.
Let $x=y=z=\frac{1}{3}$. Then $k \geqslant \frac{1}{6}$.
Below is the proof:
$$
\frac{x^{2} y^{2}}{1-z}+\frac{y^{2} z^{2}}{1-x}+\frac{z^{2} x^{2}}{1-y} \leqslant \frac{1}{6}-3 x y z \text {. }
$$

Given $x>0, y>0, z>0$ and $x+y+z=1$, we know that inequality (1) is equivalent to
$$
\begin{array}{l}
\frac{x y}{z(x+y)}+\frac{y z}{x(y+z)}+\frac{z x}{y(z+x)}+3 \leqslant \frac{1}{6 x y z} \\
\Leftrightarrow \frac{x y z}{z(x+y)}+\frac{x y z}{x(y+z)}+\frac{x y z}{y(x+z)} \leqslant \frac{1}{6(x y+y z+z x)} \\
\Leftrightarrow \frac{x y}{x+y}+\frac{y z}{y+z}+\frac{z x}{z+x} \leqslant \frac{1}{6(x y+y z+z x)} .
\end{array}
$$

From $1=(x+y+z)^{2}$
$$
\begin{array}{l}
=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x \\
\geqslant 3(x y+y z+z x) \\
\Rightarrow \frac{1}{6(x y+y z+z x)} \geqslant \frac{1}{2} .
\end{array}
$$

Also, $x y \leqslant \frac{1}{4}(x+y)^{2}$
$$
\begin{aligned}
\Rightarrow & \sum \frac{x y}{x+y} \leqslant \frac{1}{4} \sum(x+y) \\
& =\frac{1}{4}(2 x+2 y+2 z)=\frac{1}{2} .
\end{aligned}
$$

Here, “ ” denotes the cyclic sum.
Thus, inequality (2) holds, and therefore inequality (1) holds.
The equality holds if and only if $x=y=z=\frac{1}{3}$.
Hence, the minimum value of $k$ is $\frac{1}{6}$.