Solve (4y + yx^2)dy - (2x + xy^2)dx = 0 by separation of variables

If you want what dy/dx is then...

(4y + yx^2)dy = (2x + xy^2)dx

Divide by dx

(4y + yx^2)dy/dx = (2x + xy^2)

Divide by the first part

dy/dx = (2x + xy^2)/(4y + yx^2)

Simplify

dy/dx = x(2 + y^2)/y(4 + x^2)

I don' think this can be done with separation of variables.

(4y + yx^2) dy - (2x + xy^2) dx = 0
4y dy + yx^2 dy - 2x dx - xy^2 dx = 0
4y dy - 2x dx + (x^2y dy - xy^2 dx) = 0
4y dy - 2x dx + (1/2) (2x^2y dy - 2xy^2 dx) = 0
d(2y^2) - d(x^2) + (1/2) (x^2 y^2) = 0
2y^2 - x^2 + (1/2) x^2y^2 = C

(4y + yx^2) dy - (2x + xy^2) dx = 0

(4y + yx^2) dy = (2x + xy^2) dx

y(4 + x^2) dy = x(2 + y^2) dx

y/(y^2 + 2) dy = x/(x^2 + 4) dx

Integrating both sides:

1/2*ln(y^2 + 2) = 1/2*ln(x^2 + 4) + C

ln(y^2 + 2) = ln(x^2 + 4) + C

y^2 + 2 = C*(x^2 + 4)

y = +/-√[C*(x^2 + 4) - 2]

No "please" = no answer.