3. $4=8x^2+\frac{1}{4x^2}+y^2=4x^2+\frac{1}{4x^2}+4x^2+y^2 \ge 2\sqrt{4x^2.\frac{1}{4x^2}}+2\sqrt{4x^2.y^2} = 2 +4|xy|$
$\Rightarrow |xy| \le \frac12\Rightarrow -\frac12 \le x y \le \frac12.$
Vậy $\min xy = -\frac12\Leftrightarrow (x,y) = (\pm1/2, \mp1).$