6. Let $x, y, z \geqslant 0$, and $x y+y z+z x=1$, (1) find the maximum value of $f(x, y, z)=x(1-$ $\left.y^{2}\right)\left(1-z^{2}\right)+y\left(1-z^{2}\right)\left(1-x^{2}\right)+z\left(1-x^{2}\right)\left(1-y^{2}\right)$; (2) find the minimum value of $\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}$.
6. (1) By performing an identity transformation (and conditions) on $f(x, y, z)$, we have: $f(x, y, z)=4 x y z$. Using the arithmetic mean inequality, we have $x y z \leqslant\left(\frac{1}{3}\right)^{\frac{3}{2}}$, hence
$f(x, y, z)=4 x y z \leqslant \frac{4 \sqrt{3}}{9}$, when $x=y=z=\frac{\sqrt{3}}{3}$, the inequality achieves equality.
(2) Using $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$ (easy to prove), and $(x y+y z+z x)\left(\frac{1}{x y}+\frac{1}{y z}+\right.$ $\left.\frac{1}{x z}\right) \geqslant 9$, then $\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}} \geqslant \frac{1}{x y}+\frac{1}{y z}+\frac{1}{x z} \geqslant \frac{9}{x y+y z+z x}=9$, when and only when $x=y$ $=z=\frac{\sqrt{3}}{3}$, the inequality achieves equality, hence the minimum value of $\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}$ is 9.