detailed solution

Correct option is š

For the curve $x^3-3x{y}^2+2=0$, the gradient is $\frac{dy}{dx}=-\frac{3x^2-3y^2}{-6xy}=\frac{x^2-y^2}{2xy}$

For the curve $3x^{2}y-y^3-2=0$, the gradient is $\frac{dy}{dx}=-\frac{6xy}{3x^2-3y^2}=-\frac{2xy}{x^2-y^2}$

Observing the above two gradients, the product of slopes is -1

hence, the angle between the curves is ${90}^0$

detailed solution

Correct answer is 90

For the curve $x^3-3x{y}^2+2=0$, the gradient is $\frac{dy}{dx}=-\frac{3x^2-3y^2}{-6xy}=\frac{x^2-y^2}{2xy}$

For the curve $3x^{2}y-y^3-2=0$, the gradient is $\frac{dy}{dx}=-\frac{6xy}{3x^2-3y^2}=-\frac{2xy}{x^2-y^2}$

Observing the above two gradients, the product of slopes is -1

hence, the angle between the curves is ${90}^0$