1. **State the problem:** Given the curve defined by the equation $$x^3 + 3x^2y - y^3 = 3,$$ we need to: (a) Show that $$\frac{dy}{dx} = \frac{x^2 + 2xy}{y^2 - x^2}.$$ (b) Find the coordinates of points where the tangent to the curve is parallel to the x-axis. 2. **Part (a): Differentiate implicitly to find $$\frac{dy}{dx}$$** - Differentiate both sides of the equation with respect to $$x$$: $$\frac{d}{dx}(x^3) + \frac{d}{dx}(3x^2y) - \frac{d}{dx}(y^3) = \frac{d}{dx}(3).$$ - Compute each derivative: - $$\frac{d}{dx}(x^3) = 3x^2$$ - For $$3x^2y$$, use product rule: $$\frac{d}{dx}(3x^2y) = 3(2xy + x^2 \frac{dy}{dx}) = 6xy + 3x^2 \frac{dy}{dx}$$ - For $$y^3$$, use chain rule: $$\frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}$$ - Substitute these into the differentiated equation: $$3x^2 + 6xy + 3x^2 \frac{dy}{dx} - 3y^2 \frac{dy}{dx} = 0.$$ - Rearrange terms to isolate $$\frac{dy}{dx}$$: $$3x^2 + 6xy = 3y^2 \frac{dy}{dx} - 3x^2 \frac{dy}{dx} = 3(y^2 - x^2) \frac{dy}{dx}.$$ - Divide both sides by $$3(y^2 - x^2)$$: $$\frac{dy}{dx} = \frac{3x^2 + 6xy}{3(y^2 - x^2)} = \frac{x^2 + 2xy}{y^2 - x^2}.$$ 3. **Part (b): Find points where tangent is parallel to x-axis** - Tangent parallel to x-axis means slope $$\frac{dy}{dx} = 0$$. - Set numerator of $$\frac{dy}{dx}$$ equal to zero: $$x^2 + 2xy = 0.$$ - Factor: $$x(x + 2y) = 0,$$ so either $$x=0$$ or $$x + 2y=0$$. 4. **Case 1: $$x=0$$** - Substitute into original equation: $$0^3 + 3 \cdot 0^2 \cdot y - y^3 = 3 \implies -y^3 = 3 \implies y^3 = -3 \implies y = -\sqrt[3]{3}.$$ - Point: $$(0, -\sqrt[3]{3}).$$ 5. **Case 2: $$x + 2y = 0 \Rightarrow y = -\frac{x}{2}$$** - Substitute into original equation: $$x^3 + 3x^2 \left(-\frac{x}{2}\right) - \left(-\frac{x}{2}\right)^3 = 3.$$ - Simplify: $$x^3 - \frac{3}{2} x^3 - \left(-\frac{x^3}{8}\right) = 3,$$ $$x^3 - \frac{3}{2} x^3 + \frac{x^3}{8} = 3,$$ $$\left(1 - \frac{3}{2} + \frac{1}{8}\right) x^3 = 3,$$ $$\left(\frac{8}{8} - \frac{12}{8} + \frac{1}{8}\right) x^3 = 3,$$ $$\left(-\frac{3}{8}\right) x^3 = 3,$$ $$x^3 = -8,$$ $$x = -2.$$ - Find $$y$$: $$y = -\frac{-2}{2} = 1.$$ - Point: $$(-2, 1).$$ **Final answers:** - (a) $$\frac{dy}{dx} = \frac{x^2 + 2xy}{y^2 - x^2}.$$ - (b) Points where tangent is horizontal: $$(0, -\sqrt[3]{3})$$ and $$(-2, 1).$$