Ta có: $(x+y)^2\ge0 \Leftrightarrow xy\ge-\dfrac{x^2+y^2}{2}$
Từ đó suy ra: $x^2+y^2=4+xy\ge4-\dfrac{x^2+y^2}{2} \Rightarrow x^2+y^2\ge\dfrac{8}{3}$
$\min P=\dfrac{8}{3} \Leftrightarrow \left[\begin{array}{l}x=\dfrac{2}{\sqrt3};y=-\dfrac{2}{\sqrt3}\\x=-\dfrac{2}{\sqrt3};y=\dfrac{2}{\sqrt3}\end{array}\right.$
Ta có: $(x-y)^2\ge0 \Leftrightarrow xy\le\dfrac{x^2+y^2}{2}$
Từ đó suy ra: $x^2+y^2=4+xy\le4+\dfrac{x^2+y^2}{2} \Rightarrow x^2+y^2\le8$
$\max P=8 \Leftrightarrow \left[\begin{array}{l}x=y=2\\x=y=-2\end{array}\right.$