Example 2 Try to find the minimum value of the function $f(x, y)=6\left(x^{2}+y^{2}\right) \cdot (x+y)-4\left(x^{2}+x y+y^{2}\right)-3(x+y)+5$ in the region $A=\{(x, y) \mid x>0, y>0\}$.
Explanation: If $x+y \leqslant 1$, then
$$
\begin{array}{l}
x y \leqslant \frac{1}{4}(x+y)^{2} \leqslant \frac{1}{4} \text {. } \\
\text { Hence } f(x, y) \\
=6\left(x^{3}+y^{3}\right)+6 x y(x+y)-4 x y- \\
4\left(x^{2}+y^{2}\right)-3(x+y)+5 \\
=6\left(x y-\frac{1}{4}\right)(x+y-1)+(6 x+1) \text {. } \\
\left(x-\frac{1}{2}\right)^{2}+(6 y+1)\left(y-\frac{1}{2}\right)^{2}+ \\
(x+y-1)^{2}+2 \\
\geqslant 2 \text {. } \\
\end{array}
$$

When and only when $x=y=\frac{1}{2}$, the equality holds.
If $x+y>1$, then
$$
x^{2}+y^{2} \geqslant \frac{1}{2}(x+y)^{2}>\frac{1}{2} \text {. }
$$

Hence $f(x, y)$
$$
=6\left(x^{2}+y^{2}-\frac{1}{2}\right)(x+y-1)+2(x-y)^{2}+2
$$
$>2$.
In summary, for any $(x, y) \in A$, we have $f(x, y) \geqslant 2$.
When and only when $x=y=\frac{1}{2}$, $f(x, y)_{\min }=2$.