To avoid #-ve#signs, let us rewrite the eqn. as,
#y^2x+ye^(xy)=xy+1.#
Diff.ing term-wise, using the usual Rules of Diffn., we get,
#{y^2d/dx(x)+xd/dx(y^2)}+{yd/dxe^(xy)+e^(xy)d/dx(y)}#
#=d/dx(xy)+0.#
#:. y^2(1)+x{d/dy(y^2)d/dx(y)}+ye^(xy)d/dx(xy)+e^(xy)dy/dx=d/dx(xy).#
#:. y^2+x{(2y)dy/dx}+ye^(xy){xdy/dx+yd/dx(x)}+e^(xy)dy/dx={xdy/dx+yd/dx(x)}.#
#:. y^2+2xydy/dx+ye^(xy){xdy/dx+y}+e^(xy)dy/dx=xdy/dx+y.#
#;. {2xy+xye^(xy)+e^(xy)-x}dy/dx=y-y^2-y^2e^(xy).#
#:. dy/dx={y(1-y-ye^(xy))}/{2xy+xye^(xy)+e^(xy)-x}.#
This Answer is fairly reasonable, but, if we replace #e^(xy)# from this, we can have better Answer as shown below :
#NR.=y{1-y-(xy+1-y^2x)}=y(1-y-xy-1+y^2x)=y(xy^2-xy-y)=y^2(xy-x-1).#
#DR.=2xy-x+x(xy+1-y^2x)+(xy+1-y^2x)/y,#
#=(2xy^2-xy+x^2y^2+xy-y^3x^2+xy+1-y^2x)/y#
#=(xy+xy^2+x^2y^2-x^2y^3+1)/y.#
#rArr dy/dx={y^3(xy-x-1)}/(xy+xy^2+x^2y^2-x^2y^3+1).#
