II. (40 points) Given positive real numbers $x, y, z$ satisfying
$$
(x+y+z)^{3}=32 x y z \text {. }
$$

Find the maximum and minimum values of $f(x, y, z)=\frac{x^{4}+y^{4}+z^{4}}{(x+y+z)^{4}}$.
Let's assume $x+y+z=4$. Then $xyz=2$.
Therefore, we only need to find the extremum of $\frac{x^{4}+y^{4}+z^{4}}{4^{4}}$. At this point, we have
$$
\begin{array}{l}
x^{4}+y^{4}+z^{4} \\
=\left(x^{2}+y^{2}+z^{2}\right)^{2}-2\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right) \\
=[16-2(x y+y z+z x)]^{2}- \\
2(x y+y z+z x)^{2}+4 x y z(x+y+z) \\
=2 a^{2}-64 a+288 \text {, } \\
\end{array}
$$

where $a=x y+y z+z x$.
From $y+z=4-x$ and $yz=\frac{2}{x}$, we know
$$
(4-x)^{2} \geqslant \frac{8}{x} \text {. }
$$

Solving this, we get $3-\sqrt{5} \leqslant x \leqslant 2$.
Similarly, $y, z \in[3-\sqrt{5}, 2]$.
This indicates that
$$
(x-2)(y-2)(z-2) \leqslant 0 \text {, }
$$

and $(x-3+\sqrt{5})(y-3+\sqrt{5})(z-3+\sqrt{5}) \geqslant 0$.
From this, we solve that $a \in\left[5, \frac{5 \sqrt{5}-1}{2}\right]$.
And $\frac{x^{4}+y^{4}+z^{4}}{4^{4}}=\frac{(a-16)^{2}-112}{128}$, so
$$
f(x, y, z)=\frac{x^{4}+y^{4}+z^{4}}{(x+y+z)^{4}}
$$

has a minimum value and a maximum value of $\frac{383-165 \sqrt{5}}{256}$ and $\frac{9}{128}$, respectively, and these are achieved when $(x, y, z)=\left(3-\sqrt{5}, \frac{1+\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$ and $(2,1,1)$.