This has the form of an exact DE. Let's see if it is.
First rewrite: (e^x+y+ye^y)+(xe^y-1)dy/dx=0=A(x,y)+B(x,y)dy/dx=0 where A=e^x+y+ye^y and B=xe^y-1.
(p.d.=partially differentiate; wrt=with respect to)
Take A and p.d. wrt y=1+ye^y+e^y (a)
Take B and p.d wrt x=e^y (b)
Since (a)<>(b) we don't have an exact DE. (If it had been an exact DE, the two quantities would have been equal.)
Can we find an integrating factor u such that p.d. wrt y of uA = p.d. wrt x of uB?
That is, can we find u for (at this point u can be a function of x or y or both):
D(ue^x)/Dy+D(uy)/Dy+D(uye^y)/Dy=D(uxe^y)/Dx-Du/Dx? (D is del or p.d.)
e^xDu/Dy+u+yDu/Dy+u(ye^y+e^y)+ye^yDu/Dy=ue^y+xe^yDu/Dx-Du/Dx.
This simplifies a little: e^xDu/Dy+u+yDu/Dy+uye^y+ye^yDu/Dy=xe^yDu/Dx-Du/Dx.
u(1+ye^y)=(xe^y-1)Du/Dx-(e^x+y+ye^y)Du/Dy.
If u is stipulated to be a function of x only then Du/Dy=0 and the above simplifies further:
u(1+ye^y)=(xe^y-1)Du/Dx.
Similarly, if u is a function of y only then Du/Dx=0 and:
u(1+ye^y)=-(e^x+y+ye^y)Du/Dy.
More to follow...