Just Cauchy it

$(x+y)^2 \ge 0 => x^2+y^2 \ge -2xy => 3(x^2+y^2) \ge 2(x^2+y^2-xy) => x^2+y^2 \ge 2$

Equality holds when $x=1,y=-1$ or $x=-1,y=1$ .

$(x-y)^2 \ge 0 => x^2+y^2-2xy \ge 0 => 2(x^2+y^2-xy) \ge x^2+y^2 => x^2+y^2 \le 6$

Equality holds when $x=y=\sqrt{3}$ .

Transform the problem into polar coordinates: $x = r \cos \theta \\ y = r \sin \theta$

Then the condition becomes $r^2(1 - \cos \theta \sin \theta) = 3$ and we have to find the extreme values of $r^2$ . From the condition we know that $r^2 = \frac{3}{1 - \cos \theta \sin \theta} = \frac{3}{1 - \frac{\sin 2 \theta}{2} } = \frac{6}{2 - \sin 2 \theta}$ . Now let's just do the substitution $\phi = 2 \theta$ to get $r^2 = \frac{6}{2 - \sin \phi}$ .

We know that the sine function oscillates between 1 and -1. When $\sin \phi = 1$ we get the maximum $r^2 = \frac{6}{2 - 1} = 6$ . When $\sin \phi = -1$ we get the minimum $r^2 = \frac{6}{2 + 1} = 2$ .

The solution is $2 + 6 = 8$ .

As we can see, $2x^2-2xy+2y^2=6 \Rightarrow (x-y)^2+x^2+y^2=6\\$

Let $\ t=x-y$

Then $\quad$ $\ x^2+y^2=6-t^2$

So we need to find the range of $\mathit{t}\$

Obviously $\ x=t+y$

So $\ (t+y)^2-(t+y)y+y^2=3 \Rightarrow y^2+ty+t^2-3=0\$ Regard $\ \mathit{y}\$ as the pivot element.

the discriminant $\Delta = t^2-4(t^2-3) \geqslant 0 \Rightarrow -2\leqslant t \leqslant 2$

Therefore,

when $\ t=2\ or\ t=-2\ \ \ (x^2+y^2)_{min}=2$

when $\ t=0\ \ \ \ (x^2+y^2)_{max}=6$

So we can draw the conclusion that $(x^2+y^2)_{min}+(x^2+y^2)_{max}=8$