Knowing that, ln(e^(x-y^3))=(x-y^3)lne=x-y^3, we have,
-2=y(x-y^3)+xe^(x-y), or, xe^(x-y)=y(y^3-x)-2, i.e.,
xe^(x-y)=y^4-xy-2.
:. d/dx{xe^(x-y)}=d/dx{y^4-xy-2}.
Using the Product and Sum/Difference Rule, we get,
xd/dx{e^(x-y)}+e^(x-y)d/dx{x}=d/dx(y^4)-d/dx(xy)-d/dx(2), or,
xd/dx{e^(x-y)}+e^(x-y)=d/dx(y^4)-{xd/dx(y)+yd/dx(x)}....(1)
Here, by the Chain Rule, d/dx{e^(x-y)}=e^(x-y)d/dx(x-y),
=e^(x-y){d/dx(x)-d/dx(y)}=e^(x-y)(1-dy/dx).................(2).
Also, d/dx(y^4)=d/dy(y^4)*dy/dx=4y^3dy/dx....................................(3).
Utilising (2) and (3) in (1), we get,
xe^(x-y)(1-dy/dx)+e^(x-y)=4y^3dy/dx-xdy/dx-y,
:. xe^(x-y)-xe^(x-y)dy/dx+e^(x-y)=(4y^3-x)dy/dx-y,
:. xe^(x-y)+e^(x-y)+y=xe^(x-y)dy/dx+(4y^3-x)dy/dx,
={xe^(x-y)+4y^3-x}dy/dx.
:. dy/dx={xe^(x-y)+e^(x-y)+y}/{xe^(x-y)+4y^3-x}.
This Answer is quite fair, but it can be put as :
Using, xe^(x-y)=y^4-xy-2, &, e^(x-y)=(y^4-xy-2)/x,
dy/dx={y^4-xy-2+(y^4-xy-2)/x+y}/{y^4-xy-2+4y^3-x}.
={x(y^4-xy-2)+(y^4-xy-2)+xy}/{x(y^4-xy-2+4y^3-x)},
rArr dy/dx={x(y^4-xy-2)+y^4-2}/{x(y^4-xy-2+4y^3-x)}.
Enjoy Maths.!
