Find the complete solution of xdy+(x^3+xy^2-y)dx=0

Hi there! I am not so sure if this is the right answer but I will show it:

First we factor out [tex]xd[/tex] from the equation and get:
[tex]xd(y+ x^{3} +x y^{2} -y)=0 \\ xd(x^3+xy^2)=0 \\ x^{2} d(x^2+y^2)=0[/tex]

So our 1st case is that [tex]x^2d=0[/tex]
Case 1.1 [tex]x^2=0 \\ x=0[/tex]
Case 1.2 [tex]d=0[/tex]

Then our second case would be [tex] x^{2} +y^2=0[/tex]
[tex]x^2=-y^2[/tex]

Remember that all square numbers are greater than or equal to zero therefore [tex]y^2[/tex] cannot be negative and cannot be positive because that would make [tex] x^{2} [/tex] negative so [tex]y^2=0[/tex] therefore:
[tex]x^2=-y^2 \\ x=y=0[/tex]

So our solution set:
x={0}
y={0}
d={0}