Calling
N(x,y) = 2 x^4 - x^3 y + x^2 y^2 + 4 x y^3 - 4 y^4N(x,y)=2x4−x3y+x2y2+4xy3−4y4
D(x,y)=2 x^2 + x y - 2 y^2D(x,y)=2x2+xy−2y2
Dividing N(x,y)N(x,y) into D(x,y)D(x,y) will give a result Q(x,y)Q(x,y) with maximum xx coefficient equal 22 and maximum yy coefficient equal 22.
The remainder R(x,y) R(x,y) will have maximum xx coefficient equal 11 and maximum yy coefficient equal 11. So
Q(x,y) = sum_{i=0,j=0}^{i=2,j=2}q_{ij}x^i y^jQ(x,y)=i=2,j=2∑i=0,j=0qijxiyj
and
R(x,y) = r_{11}x y + r_{10}x+r_{01}y+r_{00}R(x,y)=r11xy+r10x+r01y+r00
Expanding
N(x,y)=D(x,y)Q(x,y)+R(x,y)N(x,y)=D(x,y)Q(x,y)+R(x,y)
and equating the coefficients we have (I am considering only nontrivial relationships)
{
( -4 + 2 q_{02}=0), (-q_{00} - r_{11}=0),( -q_{01} + 2 q_{10}=0),
(4 - q_{02} + 2 q_{11}=0), (-2 q_{01} - q_{10}=0),
(1 - 2 q_{02} - q_{11} + 2 q_{20}=0), (-q_{12} + 2 q_{21}=0),
(-1 - 2 q_{11} - q_{20}=0), (-2 q_{12} - q_{21}=0),
(2 - 2 q_{20}=0)
:}
solving we have
q_{02}=1,q_{11}=-1,q_{20}=1 and all others nulls. So the result is
(N(x,y))/(D(x,y))=x^2 - x y + 2 y^2
