The two curves; x³ – 3xy² + 2 = 0 and 3x²y – y³ – 2 = 0 intersect at an angle of

The correct option is π/4.

To analyze the intersection of the two curves, we first denote them as:
Curve 1: $x^3-3x{y}^2+2=0$
Curve 2: $3x^{2}y-y^3-2=0$

First, we find the slope of Curve 1 by differentiating implicitly with respect to $x$:
$3x^2-3\left(y^2+2xy\frac{dy}{dx}\right)=0$
Dividing by 3:
$x^2-y^2-2xy\frac{dy}{dx}=0$
Solving for the slope $m_1$:
$m_1=\frac{dy}{dx}=\frac{x^2-y^2}{2xy}$

Next, we differentiate Curve 2 implicitly with respect to $x$:
$3\left(2xy+x^2\frac{dy}{dx}\right)-3y^2\frac{dy}{dx}=0$
Dividing by 3:
$2xy+\left(x^2-y^2\right)\frac{dy}{dx}=0$
Solving for the slope $m_2$:
$m_2=\frac{dy}{dx}=\frac{-2xy}{x^2-y^2}$

Taking the product of the two slopes:
$m_1\cdot m_2=\left(\frac{x^2-y^2}{2xy}\right)\cdot \left(\frac{-2xy}{x^2-y^2}\right)=-1$
As the product of the slopes equals $-1$, the curves are orthogonal, corresponding to the provided correct option of π/4.