Question:

Step 1: Find the slope of the tangent to the first curve by implicit differentiation. For the curve $x^3 - 3xy^2 + 2 = 0$, we differentiate both sides with respect to $x$: $$3x^2 - 3y^2 - 6xy\frac{dy}{dx} = 0$$ Solving for $\frac{dy}{dx}$: $$6xy\frac{dy}{dx} = 3x^2 - 3y^2$$ $$\frac{dy}{dx} = \frac{3x^2 - 3y^2}{6xy} = \frac{x^2 - y^2}{2xy}$$ Let $m_1 = \dfrac{x^2 - y^2}{2xy}$ be the slope of the tangent to the first curve. Step 2: Find the slope of the tangent to the second curve by implicit differentiation. For the curve $3x^2y - y^3 - 2 = 0$, we differentiate both sides with respect to $x$: $$6xy + 3x^2\frac{dy}{dx} - 3y^2\frac{dy}{dx} = 0$$ Solving for $\frac{dy}{dx}$: $$3x^2\frac{dy}{dx} - 3y^2\frac{dy}{dx} = -6xy$$ $$(3x^2 - 3y^2)\frac{dy}{dx} = -6xy$$ $$\frac{dy}{dx} = \frac{-6xy}{3x^2 - 3y^2} = \frac{-2xy}{x^2 - y^2}$$ Let $m_2 = \dfrac{-2xy}{x^2 - y^2}$ be the slope of the tangent to the second curve. Step 3: Determine the relationship between the two slopes. We compute the product of the two slopes: $$m_1 \cdot m_2 = \frac{x^2 - y^2}{2xy} \cdot \frac{-2xy}{x^2 - y^2}$$ $$m_1 \cdot m_2 = \frac{(x^2 - y^2) \cdot (-2xy)}{2xy \cdot (x^2 - y^2)}$$ $$m_1 \cdot m_2 = -1$$ Step 4: Conclude the angle between the curves. When the product of the slopes of two curves equals $-1$, the tangent lines are perpendicular to each other. This means the curves intersect at a right angle at every intersection point. Therefore, the angle between the curves is $\boxed{\dfrac{\pi}{2}}$. The correct answer is **Option 1**.