i am not good at solving these problems.
If it is with respect to x, or if you want dy/dx or y', then,
[x(2y dy/dx) +(y^2)(1 dx/dx)] +[y(2x dx/dx) +(x^2)(1 dy/dx)] = 0
You see, "divide everything" by dx if it is with rexpect to x.
[(2xy dy/dx ) +(y^2)] +[(2xy) +(x^2 dy/dx)] = 0
It does not have to be as slow as that. It's just mny way of explaining the basics or the details to you.
(2xy)y' +y^2 +2xy +(x^2)y' = 0
Collect the (y')s,
(2xy +x^2)y' = -2xy -y^2
x(2y +x)y' = -y(2x +y)
y' = -y(2x +y) / x(x +2y) -----------answer.