II. (40 points) Let real numbers $x, y, z \in (0, \sqrt{5})$ satisfy $x^{4} + y^{4} + z^{4} \geqslant 27$.
Find the maximum value of $\frac{x}{x^{2}-5} + \frac{y}{y^{2}-5} + \frac{z}{z^{2}-5}$.
Let
$$
f(m)=\frac{2 \sqrt{6}}{9} m^{3}\left(5-m^{2}\right)(0<m<\sqrt{5}) .
$$

Then \( f^{2}(m)=\frac{2}{3} m^{2} \cdot \frac{2}{3} m^{2} \cdot \frac{2}{3} m^{2}\left(5-m^{2}\right)\left(5-m^{2}\right) \)
$$
\begin{aligned}
\leqslant & \left\{\frac { 1 } { 5 } \left[\frac{2}{3} m^{2}+\frac{2}{3} m^{2}+\frac{2}{3} m^{2}+\left(5-m^{2}\right)+\right.\right. \\
& \left.\left.\left(5-m^{2}\right)\right]\right\}^{5} \\
= & 2^{5} .
\end{aligned}
$$

Thus, \( f(m) \leqslant 4 \sqrt{2} \).
Therefore, \( 0 < m^{3}\left(5-m^{2}\right) \leqslant 6 \sqrt{3} \).
Equality holds if and only if \( m=\sqrt{3} \).
$$
\begin{array}{l}
\text { Hence } \frac{x}{5-x^{2}}+\frac{y}{5-y^{2}}+\frac{z}{5-z^{2}} \\
=\frac{x^{4}}{x^{3}\left(5-x^{2}\right)}+\frac{y^{4}}{y^{3}\left(5-y^{2}\right)}+\frac{z^{4}}{z^{3}\left(5-z^{2}\right)} \\
\geqslant \frac{x^{4}}{6 \sqrt{3}}+\frac{y^{4}}{6 \sqrt{3}}+\frac{z^{4}}{6 \sqrt{3}} \\
=\frac{1}{6 \sqrt{3}}\left(x^{4}+y^{4}+z^{4}\right) \geqslant \frac{27}{6 \sqrt{3}}=\frac{3 \sqrt{3}}{2} .
\end{array}
$$

Thus, \( \frac{x}{x^{2}-5}+\frac{y}{y^{2}-5}+\frac{z}{z^{2}-5} \leqslant-\frac{3 \sqrt{3}}{2} \).
Equality holds if and only if \( x=y=z=\sqrt{3} \).
Therefore, the maximum value is \( -\frac{3 \sqrt{3}}{2} \).