We rewrite the given eqn. as #1+e^(xy)=y^2(x+1)^2#.
Diff.ing both sides w.r.t. #x#, we get,
#(1+e^(xy))'={y^2(x+1)^2}'#.
#:. 1'+{e^(xy)}'=(y^2){(x+1)^2}'+(x+1)^2(y^2)'#.
#:. 0+{e^(xy)}'=y^2{2(x+1)(x+1)'}+(x+1)^2(2yy')#.
#:. e^(xy)(xy)'=2y^2(x+1)(x'+1')+2yy'(x+1)^2#.
#:. e^(xy){xy'+yx'}=2y^2(x+1)(1+0)+2yy'(x+1)^2#.
#:. (xy'+y)e^(xy)=2y^2(x+1)+2yy'(x+1)^2#
#:. xy'e^(xy)-2yy'(x+1)^2=2y^2(x+1)-ye^(xy)#
#:. y'{xe^(xy)-2y(x+1)^2}=y{2y(x+1)-e^(xy)}#
#:. y'=[y{2y(x+1)-e^(xy)}]/{xe^(xy)-2y(x+1)^2}#
This can further be simplified by replacing #e^(xy)# by #y^2(x+1)^2-1#
#:. y'=[y{2y(x+1)-y^2(x+1)^2+1}]/{xy^2(x+1)^2-x-2y(x+1)^2}#
Enjoy Maths.!
