Find the orthogonal trajectories of the following family of curves: $x^3y - xy^3 = \alpha$ The answer is $-6x^2y^2 + x^4 + y^4 = \beta$, but I get $-3x^2y^2 + \frac{x^4}{4} + \frac{y^4}{4} = \beta$ and I can't seem to find my mistake. My work:
Differentiaing the given equation, we get 
$$3x^2y + x^3\frac{dy}{dx} - y^3 - 3y^2x\frac{dy}{dx} = 0. $$
So $$3x^2y - y^3 = (3y^2x -x^3)\frac{dy}{dx}, $$ and hence 
$$ \frac{dy}{dx} = \frac{3x^2y - y^3}{3y^2x - x^3}. $$
Now, the slope of the other family of curves will be the negative inverse, so for the orthogonal trajectory, we have $$\frac{dy}{dx} = -\frac{3y^2x-x^3}{3x^2y-y^3}, $$ which we can write as 
$$-(3y^2x-x^3)dx = (3x^2y-y^3)dy. $$ Integration of the last differential equation yields 
$$-\frac{3}{2}y^2x^2 + \frac{x^4}{4} = \frac{3}{2}x^2y^2 - \frac{y^4}{4} + \beta, $$ $\beta$ being an arbitrary constant. 
This simplifies to $$-3y^2x^2 + \frac{x^4}{4} + \frac{y^4}{4} = \beta. $$ which is not correct.
Where am I going wrong?

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.
