x = 2/3
y = 1/2
\frac{(2y^2-3x)^2-(2y-x)(2y+x)*y+12xy^2+4y^3}{(4x+y)^2+(x-2y)(x+2y)*4-8xy}=
\frac{4y^4-12xy^2+9x^2-(4y^2-x^2)y+12xy^2+4y^3}{16x^2+8xy+y^2+(x^2-4y^2)*4-8xy}=
\frac{4y^4+9x^2-(4y^3-x^2y)+4y^3}{16x^2+y^2+4x^2-16y^2}=
\frac{4y^4+9x^2-4y^3+x^2y+4y^3}{20x^2-15y^2}=\frac{4y^4+9x^2+x^2y}{20x^2-15y^2}=
\frac{4*(\frac{1}{2})^4+9*(\frac{2}{3})^2+(\frac{2}{3})^2*\frac{1}{2}}{20*(\frac{2}{3})^2-15*(\frac{1}{2})^2}=
\frac{4*\frac{1}{16}+9*\frac{4}{9}+\frac{4}{9}*\frac{1}{2}}{20*\frac{4}{9}-15*\frac{1}{4}}=\frac{\frac{1}{4}+\frac{36}{9}+\frac{2}{9}}{\frac{80}{9}-\frac{15}{4}}=
\frac{\frac{9+144+8}{36}}{\frac{320-135}{36}}=\frac{161}{36}*\frac{36}{185}=\frac{161}{185}