$\text{ The two curves }{\mathrm{x}}^t-3{\mathrm{xy}}^2+2=0\text{ and }3{\mathrm{x}}^2\mathrm{y}-{\mathrm{y}}^3=2$

$\text{ Two curves cuts at right angle if product of their slopes }$$\text{ is }-1\text{ . }$

$\text{ Two given curves are }$

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

$\text{ and }3x^{2}y-y^3-2=0---------\left(2\right)$

$\text{ Differentiate equ. (i), }$

$\Rightarrow m_1=\frac{dy}{dx}=\frac{3\left(x^2-y^2\right)}{6xy}$

$\text{ Differentiate equ. (ii), }$

$\Rightarrow {\mathrm{m}}_2=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2\mathrm{xy}}{\left({\mathrm{x}}^2-{\mathrm{y}}^2\right)}$

$\therefore {\mathrm{m}}_1\times {\mathrm{m}}_2=\frac{\left({\mathrm{x}}^2-{\mathrm{y}}^2\right)}{2\mathrm{xy}}\times \frac{-2\mathrm{xy}}{\left({\mathrm{x}}^2-{\mathrm{y}}^2\right)}$

$\Rightarrow {\mathrm{m}}_1\times {\mathrm{m}}_2=-1$