The answer is ½.
The easy way to solve this is to give values to x and y. Let x=1 and y=2, then z=-3 because z=-(x+y).
x2y2+x2z2+y2z2=4+9+36=49; x4+y4+z4=1+16+81=98; 49/98=½.
Now let x=2, y=3, so z=-5:
x2y2+x2z2+y2z2=361; x4+y4+z4=1+16+81=722; 361/722=½.
Now to work out the reason:
z=-(x+y), so z2=(x+y)2.
x2y2+x2z2+y2z2=x2y2+z2(x2+y2)=x2y2+(x+y)2(x2+y2)=
x2y2+(x+y)2[(x+y)2-2xy]=x2y2-2xy(x+y)2+(x+y)4=(xy-(x+y)2)2=(-x2-xy-y2)2=(x2+xy+y2)2=((x2+y2)+xy)2=(x2+y2)2+2xy(x2+y2)+x2y2.
x4+y4+z4=x4+y4+(x+y)4=x4+y4+(x2+2xy+y2)2=x4+y4+((x2+y2)+2xy)2=
x4+y4+(x2+y2)2+4xy(x2+y2)+4x2y2=(x2+y2)2-2x2y2+(x2+y2)2+4xy(x2+y2)+4x2y2=2(x2+y2)2+4xy(x2+y2)+2x2y2=
2((x2+y2)2+2xy(x2+y2)+x2y2).
Therefore:
(x2y2+x2z2+y2z2)/(x4+y4+z4)=((x2+y2)2+2xy(x2+y2)+x2y2)/(2((x2+y2)2+2xy(x2+y2)+x2y2))=½.
That is: (x2+xy+y2)2/(2(x2+xy+y2)2)=½.