#(x/(x-4y))=x^3/y^2-1=(x^3-y^2)/y^2#
#:.xy^2=(x-4y)(x^3-y^2)=x^4-xy^2-4x^3y+4y^3#
#:.xy^2+xy^2+4x^3y=x^4+4y^3#
#:.2xy^2+4x^3y=x^4+4y^3#
Diff.ing both sides w.r.t. #x#,
#d/dx(2xy^2)+d/dx(4x^3y)=d/dx(x^4)+d/dx(4y^3)#
#:. 2{xd/dx(y^2)+y^2d/dx(x)}+4{x^3dy/dx+3x^2y}#
#=4x^3+4d/dx(y^3)........(1)#
Here, let us note that, by Chain Rule,
#d/dx(y^2)=d/dy(y^2)*dy/dx=2y*dy/dx=2yy'#,
and on the same lines, #d/dx(y^3)=3y^2y'#.
These results will be used in #(1)# [after cancelling #2# throughout]
to get,
#{x*2yy'+y^2*1}+2{x^3y'+3x^2y}=2x^3+2*3y^2y'#
#2xyy'+y^2+2x^3y'+6x^2y=2x^3+6y^2y'#
#y'(2xy+2x^3-6y^2)=2x^3-y^2-6x^2y#
#y'=(2x^3-y^2-6x^2y)/(2xy+2x^3-6y^2)#
