This is a differential expression, not an equation, so I assume that the expression equals zero.
xy³dx+(y+1)e⁻ˣdy=0
Rewriting:
(y+1)e⁻ˣdy=-xy³dx
-((y+1)/y³)(dy)=xeˣdx
(-1/y²-1/y³)dy=xeˣdx
Integrating each side:
1/y+1/(2y²)=xeˣ-eˣ+C where C is the integration constant.
This can be written:
2y+1=2xy²eˣ-2y²eˣ+2Cy² or 2y²(xeˣ-eˣ+C).
It can also be written:
2y²(xeˣ-eˣ+C)-2y-1=0.
If the differential expression was a constant a (which may or may not be zero), then the solution would be:
2y²(xeˣ-eˣ+ax+C)-2y-1=0.