Prove that lines $y^3 - x^3 + 3xy(y -x) = 0$ are equally inclined to each other. 
Prove that lines $y^3 - x^3 + 3xy(y -x) = 0$ are equally inclined to each other.

I have to find this by using polar form of this equation.

Without using polar form I get,
$$(y - x)(y^2 + 4xy + x^2)= 0$$
$$\implies y = x \ or \ y^2 + 4xy + x^2 = 0$$
Solving second part,
$$y^2 + 4xy + x^2 = 0 \implies x= y(-2 \pm \sqrt{3})$$
Therefore the angles made by these three lines with the x-axis are $45^\circ, 105^\circ$ and $165^\circ$. 
Hence proved.
With polar form, 
On transforming the equation I get 
$$(\sin \theta - \cos \theta)(1 + 4\sin \theta \cos\theta) = 0$$
$$ \implies \tan\theta = 1, \sin 2\theta = -\frac12$$
$$\implies \theta = n\pi + \frac\pi4, {1 \over 2}\left(n\pi + (-1)^n\frac{7\pi}6\right) \tag{*}$$
But I don't know what to do with this now ? how can I show the lines are equally inclined with the help of (*) ?

Your calculation is correct, so just contrast them. Take $n=1,2,3$
$$\dfrac{7\pi}{12}-\dfrac{\pi}{4}=\dfrac{\pi}3$$
$$\dfrac{11\pi}{12}-\dfrac{7\pi}{12}=\dfrac{\pi}3$$.
note: $0\leq\theta\leq\pi$.
