Can you find the point of intersection?

Differentiating $x^{3}-3xy^{2}=a$ w.r.t. $x$ , we obtain :

$3x^{2}-3\left( x.2y.\dfrac{dy}{dx}+1.y^{2}\right)=0$

$\implies x^{2}-y^{2}-2.x.y.\dfrac{dy}{dx}=0$

therefore, $\dfrac{dy}{dx}=\dfrac{x^{2}-y^{2}}{2.x.y}$ _ _ $(1)$

Now, differentiating $3yx^{2}-y^{3}=b$ w.r.t. $x$ , we get :

$3 \left( 2xy+x^{2}.\dfrac{dy}{dx} \right) -3y^{2}.\dfrac{dy}{dx}=0$

$\implies (x^{2}-y^{2}).\dfrac{dy}{dx}+2.x.y=0$

therefore, $\dfrac{dy}{dx}=-\dfrac{2.x.y}{x^{2}-y^{2}}$ _ _ $(2)$

Let the two curves intersect at point P(r,s) ,

From $(1)$ at P :

$\left( \dfrac{dy}{dx} \right)_{1}=\dfrac{r^{2}-s^{2}}{2.r.s}$ for curve $(1)$

From $(2)$ at P:

$\left( \dfrac{dy}{dx} \right)_{2}=-\dfrac{2.r.s}{r^{2}-s^{2}}$ for curve $(2)$

Since $\left( \dfrac{dy}{dx} \right)_{1}$ . $\left( \dfrac{dy}{dx} \right)_{2}=-1$

Therefore the curves $x^{3}-3xy^{2}=a$ and $3yx^{2}-y^{3}=b$ cut each other at right angle.

We can say both the curves are orthogonal .

Let the point of intersection of given curves be (h, k)

$m_{1}= \frac{h^{2}-k^{2}}{2hk}$

$m_{2}= \frac{2hk}{k^{2}-h^{2}}$

$\boxed{m_{1} \times{m_{2}} =-1}$