Example 4 Try to find the minimum value of the function $f(x, y)=6\left(x^{2}+y^{2}\right)(x+y)-4\left(x^{2}+x y+y^{2}\right)-3(x+y)+5$ in the region $D=\{(x, y) \mid x>0, y>0\}$.
Notice that $x>0, y>0$, without loss of generality, let $x+y=k(k>0)$, it is obvious that
$$
x \cdot y \leqslant \frac{1}{4}(x+y)^{2}=\frac{1}{4} k^{2}, x^{2}+y^{2} \geqslant \frac{1}{2}(x+y)^{2}=\frac{1}{2} k^{2} \text {. }
$$

Consider taking $k=1$, that is, when $x+y \leqslant 1$, we have $x y \leqslant \frac{1}{4}(x+y)^{2} \leqslant \frac{1}{4}$.
$$
\begin{aligned}
f(x, y)= & 6\left(x^{3}+y^{3}\right)+6 x y(x+y)-4 x y-4\left(x^{2}+y^{2}\right)-3(x+y)+5 \\
= & 6\left(x y-\frac{1}{4}\right)(x+y-1)+6 x\left(x-\frac{1}{2}\right)^{2}+6 y\left(y-\frac{1}{2}\right)^{2}+\left(x-\frac{1}{2}\right)^{2}+ \\
& \left(y-\frac{1}{2}\right)^{2}+(x+y-1)^{2}+2 \\
= & 6\left(x y-\frac{1}{4}\right)(x+y-1)+(6 x+1)\left(x-\frac{1}{2}\right)^{2}+(6 y+1)\left(y-\frac{1}{2}\right)^{2}+ \\
& (x+y-1)^{2}+2
\end{aligned}
$$
$\geqslant 2$, where the equality holds if and only if $x=y=\frac{1}{2}$.
When $x+y>1$, we have $x^{2}+y^{2} \geqslant \frac{1}{2}(x+y)^{2}>\frac{1}{2}$.
$$
f(x, y)=6\left(x^{2}+y^{2}-\frac{1}{2}\right)(x+y-1)+2(x-y)^{2}+2>2 \text {. }
$$

In summary, for any given $(x, y) \in D$, we have $f(x, y) \geqslant 2$, and when $x=y=\frac{1}{2}$, $f(x)_{\text {min }}=2$.