3. Let $x, y$ be real numbers. Then
$$
f(x, y)=x^{2}+x y+y^{2}-x-y
$$

the minimum value is $\qquad$ .
3. $-\frac{1}{3}$.

Let $u=x+y, v=x-y$. Then $x=\frac{u+v}{2}, y=\frac{u-v}{2}$.
Thus $x^{2}+x y+y^{2}-x-y$ $=\left(\frac{u+v}{2}\right)^{2}+\frac{u+v}{2} \cdot \frac{u-v}{2}+\left(\frac{u-v}{2}\right)^{2}-u$ $=\frac{3 u^{2}-4 u+v^{2}}{4}$ $=\frac{3\left(u-\frac{2}{3}\right)^{2}}{4}+\frac{v^{2}}{4}-\frac{1}{3} \geqslant-\frac{1}{3}$.