Example 1 Find the real solutions of the equation $x+x y+y^{2}-3 x-3 y$ ~ $3=0$.

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Solution: Transform the original equation into
$$
x^{2}+(y-3) x+\left(y^{2}-3 y+3\right)=0 \text {. }
$$

Since the required $x$ is a real number, then $\Delta \geqslant 0$, i.e., $(y-3)^{2}-4\left(y^{2}-3 y+3\right) \geqslant 0$.
Rearranging, we get $(y-1)^{2} \leqslant 0$.
Since the required $y$ is also a real number, it must be that $(y-1)^{2}=$ 0, i.e., $y=1$.
Substituting $y=1$ into the original equation, we get $x=1$.
Therefore, the real solution to the original equation is $x=1, y=1$. Note: For a bivariate quadratic equation or an equation that can be transformed into a bivariate quadratic equation, transforming it into an equation about $x$ (or $y$), and using the discriminant to establish an inequality for solving is very concise.