Find the orthogonal trajectories of the following family of curves: $x^3y - xy^3 = \alpha$

Since this is tagged complex analysis,

let $z = x+iy$ and let $f(x+iy) = x^3y - xy^3$. Expressed in terms of $z$, this is $f(z) = \frac 1 {16i}\left((z+\bar{z})^3(z - \bar z) + (z + \bar z)(z - \bar z)^3\right) = \frac 1{8i}(z^4 - \bar z^4) = \frac 1 4 Im(z^4) = Im(z^4/4)$.

So the curves $x^3y-xy^3 = A$ are the pullback of the horizontal lines by the holomorphic function $z \mapsto z^4/4$. Since holomorphic functions preserve angles (except when the derivative vanish, here it is at $z=0$), the curves that are always orthogonal to those, are the pullback of the vertical lines by $z \mapsto z^4/4$, i.e. the curves $Re(z^4/4) = B$.

And finally, $Re(z^4/4) = (x^4-6x^2y^2+y^4)/4$

At $z=0$, $f$ multiplies angles by $4$ and is $4$-to $1$ around $z=0$, so the two curves $Re(f(z)) = 0$ and $Im(f(z)) = 0$ both have a star-like singular point (an $8$-branched star), and the angle between the two stars is $\pi/8$

I t seems that we have $$y'=\frac{3xy^2-x^3}{y^3-3x^2y}$$ instead, which is an homogenous OE. We have $$\frac{du}{\frac{3u^2-1}{u^3-3u^2}-u}=\frac{dx}{x}~\equiv~\frac{(-u^3+3u^2)du}{u^4+3u^2-3u^2+1}=\frac{dx}{x},~~~~~~u=y/x,x\neq 0$$ in which the right ODE is a separable one.