Two curves x^3-3xy^2+2=0 and 3x^2y-y^3=2:

The correct answer is Option (2) → Cut at right angle

$x^3-3xy^2+2=0$ and $3x^2y-y^3=2$

finding slopes

for first $3x^2-3y^2-6xy\frac{dy}{dx}=0$

$x^2-y^2=2xy\frac{dy}{dx}$

$\frac{dy}{dx}=m=\frac{x^2-y^2}{2xy}$

for second $6xy+3x^2\frac{dy}{dx}-3y^2\frac{dy}{dx}=0$

so $\frac{dy}{dx}=n=-\frac{2xy}{x^2-y^2}$

so $m×n=-1$

both cut each other at right angle