Example 7 Given
\[
\begin{aligned}
f(x, y)= & x^{3}+y^{3}+x^{2} y+x y^{2}- \\
& 3\left(x^{2}+y^{2}+x y\right)+3(x+y),
\end{aligned}
\]
and \( x, y \geqslant \frac{1}{2} \). Find the minimum value of \( f(x, y) \).
(2011, Hebei Province High School Mathematics Competition)
【Analysis】This problem involves finding the extremum of a bivariate function, with a very complex expression, making it difficult to find a breakthrough. Observing the symmetry in the expression, we can attempt the following transformation.

First, when $x \neq y$, multiply both sides of the function by $x-y$, yielding
$$
\begin{array}{l}
(x-y) f(x, y) \\
=\left(x^{4}-3 x^{3}+3 x^{2}\right)-\left(y^{4}-3 y^{3}+3 y^{2}\right) .
\end{array}
$$

At this point, it is easy to construct the function
$$
g(x)=x^{4}-3 x^{3}+3 x^{2} \text {. }
$$

Equation (1) can be rewritten as
$$
f(x, y)=\frac{g(x)-g(y)}{x-y} \text {. }
$$

Thus, $f(x, y)$ can be viewed as the slope of the line connecting two points on the graph of $g(x)$.
Furthermore, when $x=y$,
$$
f(x, y)=4 x^{3}-9 x^{2}+6 x \text {. }
$$

Therefore, we only need to find the minimum value of the derivative function $h(x)=4 x^{3}-9 x^{2}+6 x$ of $g(x)$ for $x \geqslant \frac{1}{2}$.

It is easy to find that when $x \geqslant \frac{1}{2}$, the minimum value of $h(x)=4 x^{3}-9 x^{2}+6 x$ is 1.
Thus, the minimum value of $f(x, y)$ is 1.