X²/y + y²/x =3 ---> x³+y³ =3xy (1) (x≠0; y≠0)
x+y =2 --> y=2-x подставляем в (1)
x³+(2-x)³ =3x(2-x)
x³+8-12x+6x²-x³ =6x-3x²
9x²-18x+8=0
D =324-288=36
x1=(18+6)/18=4/3=1 1/3 y1-2- 1 1/3=2/3
x2=(18-6)/18 =12/18=2/3 y2=2-2/3=1 1/3
ответ: (1 1/3; 2/3) (2/3; 1 1/2)