5. (1) Since
$$\begin{array}{c} x^{9}+y^{9}=\left(x^{3}+y^{3}\right)\left(x^{6}-x^{3} y^{3}+y^{6}\right) \\ \frac{x^{6}-x^{3} y^{3}+y^{6}}{x^{6}+x^{3} y^{3}+y^{6}}=1-\frac{2 x^{3} y^{3}}{x^{6}+x^{3} y^{3}+y^{6}} \geqslant 1-\frac{2 x^{3} y^{3}}{2 x^{3} y^{3}+x^{3} y^{3}}=\frac{1}{3} \end{array}$$

Therefore,
$$\frac{x^{9}+y^{9}}{x^{6}+x^{3} y^{3}+y^{6}} \geqslant \frac{1}{3}\left(x^{3}+y^{3}\right)$$

Similarly,
$$\begin{array}{l} \frac{y^{9}+z^{9}}{y^{6}+y^{3} z^{3}+z^{6}} \geqslant \frac{1}{3}\left(y^{3}+z^{3}\right) \\ \frac{z^{9}+x^{9}}{z^{6}+z^{3} x^{3}+x^{6}} \geqslant \frac{1}{3}\left(z^{3}+x^{3}\right) \end{array}$$

Therefore,
$$\frac{x^{9}+y^{9}}{x^{6}+x^{3} y^{3}+y^{6}}+\frac{y^{9}+z^{9}}{y^{6}+y^{3} z^{3}+z^{6}}+\frac{z^{9}+x^{9}}{z^{6}+z^{3} x^{3}+x^{6}} \geqslant \frac{2}{3}\left(x^{3}+y^{3}+z^{3}\right) \geqslant 2 x y z=2$$