The two curves $${x^3} - 3x{y^2} + 2 = 0$$     and $$3{x^2}y - {y^3} = 2$$

Solution :
Two curves cuts at right angle if product of their slopes is $$ –1.$$
Two given curves are
$$\eqalign{ & {x^3} - 3x{y^2} + 2 = 0......\left( {\text{i}} \right) \cr & {\text{and }}3{x^2}y - {y^3} - 2 = 0......\left( {{\text{ii}}} \right) \cr} $$
Differentiate equation $$\left( {\text{i}} \right),$$
$$\eqalign{ & 3{x^2} - 3\left[ {{y^2} + 2xy\frac{{dy}}{{dx}}} \right] = 0 \cr & \Rightarrow 3\left( {{x^2} - {y^2}} \right) = 6xy\frac{{dy}}{{dx}} \cr & \Rightarrow {m_1} = \frac{{dy}}{{dx}} = \frac{{3\left( {{x^2} - {y^2}} \right)}}{{6xy}} \cr} $$
Differentiate equation $$\left( {\text{ii}} \right),$$
$$\eqalign{ & 3{x^2}y - {y^3} - 2 = 0 \cr & \Rightarrow 3\left[ {{x^2}\frac{{dy}}{{dx}} + 2xy} \right] - 3{y^2}\frac{{dy}}{{dx}} = 0 \cr & \Rightarrow {x^2}\frac{{dy}}{{dx}} + 2xy - {y^2}\frac{{dy}}{{dx}} = 0 \cr & \Rightarrow \left( {{x^2} - {y^2}} \right)\frac{{dy}}{{dx}} = - 2xy \cr & \Rightarrow {m_2} = \frac{{dy}}{{dx}} = \frac{{ - 2xy}}{{\left( {{x^2} - {y^2}} \right)}} \cr & \therefore \,{m_1} \times {m_2} = \frac{{\left( {{x^2} - {y^2}} \right)}}{{2xy}} \times \frac{{ - 2xy}}{{\left( {{x^2} - {y^2}} \right)}} \cr & \Rightarrow {m_1} \times {m_2} = - 1 \cr} $$
i.e., curves cuts at right angle.