Calling
N(x,y) = 2 x^4 - x^3 y + x^2 y^2 + 4 x y^3 - 4 y^4
D(x,y)=2 x^2 + x y - 2 y^2
Dividing N(x,y) into D(x,y) will give a result Q(x,y) with maximum x coefficient equal 2 and maximum y coefficient equal 2.
The remainder R(x,y) will have maximum x coefficient equal 1 and maximum y coefficient equal 1. So
Q(x,y) = sum_{i=0,j=0}^{i=2,j=2}q_{ij}x^i y^j
and
R(x,y) = r_{11}x y + r_{10}x+r_{01}y+r_{00}
Expanding
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
