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#
