Solution: First let $x = y = z = \frac{1}{3}$, then $k \ge \frac{1}{6}$. Next we prove that:
$$\frac{x^2 y^2}{1-z} + \frac{y^2 z^2}{1-x} + \frac{z^2 x^2}{1-y} \le \frac{1}{6} - 3xyz. \quad (1)$$
Since $x > 0, y > 0, z > 0$ and $x + y + z = 1$, we know that inequality (1) is equivalent to
$$\begin{align*} & \frac{xy}{z(x+y)} + \frac{yz}{x(y+z)} + \frac{zx}{y(z+x)} + 3 \le \frac{1}{6xyz} \\ \Leftrightarrow & \frac{xyz}{z(x+y)} + \frac{xyz}{x(y+z)} + \frac{xyz}{y(z+x)} \le \frac{1}{6(xy+yz+zx)} \\ \Leftrightarrow & \frac{xy}{x+y} + \frac{yz}{y+z} + \frac{zx}{z+x} \le \frac{1}{6(xy+yz+zx)}. \tag{2} \end{align*}$$
From
$$\begin{align*} & \text{From } 1 = (x + y + z)^2 \\ &= x^2 + y^2 + z^2 + 2xy + 2yz + 2zx \\ &\ge 3(xy + yz + zx) \\ &\Rightarrow \frac{1}{6(xy + yz + zx)} \ge \frac{1}{2}. \\ & \text{Also } xy \le \frac{1}{4}(x + y)^2 \\ &\Rightarrow \Sigma \frac{xy}{x+y} \le \frac{1}{4} \Sigma (x+y) \\ &= \frac{1}{4}(2x + 2y + 2z) = \frac{1}{2}. \end{align*}$$
Here, "$\Sigma$" denotes the cyclic symmetric sum.

Thus, inequality (2) holds, so inequality (1) holds. Equality holds if and only if $x = y = z = \frac{1}{3}$. Therefore the minimum value of $k$ is $\frac{1}{6}$.