$$
\begin{array}{l}
\text { II. (40 points) Find } \\
f(x, y) \\
=\sqrt{\cos 4 x+7}+\sqrt{\cos 4 y+7}+ \\
\sqrt{\cos 4 x+\cos 4 y-8 \sin ^{2} x \cdot \sin ^{2} y+6}
\end{array}
$$

the minimum value.
II. Notice that,
$$
\begin{array}{l}
\cos 4 x+7=2 \cos ^{2} 2 x+6 \\
=2\left(2 \cos ^{2} x-1\right)^{2}+6 \\
=8\left(\cos ^{4} x-\cos ^{2} x+1\right) .
\end{array}
$$

Similarly, $\cos 4 y+7=8\left(\cos ^{4} y-\cos ^{2} y+1\right)$.
$$
\begin{array}{l}
\text { and } \cos 4 x+\cos 4 y-8 \sin ^{2} x \cdot \sin ^{2} y+6 \\
=(\cos 4 x+7)+(\cos 4 x+7)- \\
8 \sin ^{2} x \cdot \sin ^{2} y-8 \\
=8\left(\cos ^{4} x-\cos ^{2} x+1\right)+ \\
8\left(\cos ^{4} y-\cos ^{2} y+1\right)- \\
8\left(1-\cos ^{2} x\right)\left(1-\cos ^{2} y\right)-8 \\
=8\left(\cos ^{4} x+\cos ^{4} y\right)-8 \cos ^{2} x \cdot \cos ^{2} y .
\end{array}
$$

Thus, $f(x, y)$
$$
\begin{aligned}
= & \sqrt{8}\left(\sqrt{\cos ^{4} x-\cos ^{2} x+1}+\right. \\
& \sqrt{\cos ^{4} y-\cos ^{2} y+1}+ \\
& \left.\sqrt{\cos ^{4} x+\cos ^{4} y-\cos ^{2} x \cdot \cos ^{2} y}\right) .
\end{aligned}
$$

As shown in Figure 3, construct equilateral triangles $\triangle S A B$, $\triangle S B C$, and $\triangle S C D$ with side length 1. Take points $X$ and $Y$ on $S B$ and $S C$ respectively, such that
$$
\begin{array}{l}
S X=\cos ^{2} x, \\
S Y=\cos ^{2} y .
\end{array}
$$

Connect $A X$ and $A Y$.
Then $f(x, y)$
$$
=\sqrt{8}(A X+X Y+Y D) \text {. }
$$

The minimum value is the length of line segment $A S D$, i.e., when
$$
x=y=\frac{\pi}{2} \text {, } f_{\text {min }}=2 \sqrt{8}=4 \sqrt{2} \text {. }
$$