对于方程为 $x^2 - 3xy + y^3 = 8$ 的曲线，求

1. Differentiate the given equation implicitly with respect to $x$.

$\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2 - 3xy + y^3\right) = \frac{\mathrm{d}}{\mathrm{d}x}(8)$
$2x - 3\left(y + x\frac{\mathrm{d}y}{\mathrm{d}x}\right) + 3y^2\frac{\mathrm{d}y}{\mathrm{d}x} = 0$
$2x - 3y - 3x\frac{\mathrm{d}y}{\mathrm{d}x} + 3y^2\frac{\mathrm{d}y}{\mathrm{d}x} = 0$

2. Find an expression for $y$ in terms of $x$ by applying the condition for a horizontal tangent.

Rearranging the terms to isolate $\frac{\mathrm{d}y}{\mathrm{d}x}$:
$\frac{\mathrm{d}y}{\mathrm{d}x}(3y^2 - 3x) = 3y - 2x$
For the tangent to be parallel to the $x$-axis, $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$.
$3y - 2x = 0 \implies y = \frac{2}{3}x$

3. Substitute this relationship into the original equation and solve for $x$.

Substitute $y = \frac{2}{3}x$ into $x^2 - 3xy + y^3 = 8$:
$x^2 - 3x\left(\frac{2}{3}x\right) + \left(\frac{2}{3}x\right)^3 = 8$
$x^2 - 2x^2 + \frac{8}{27}x^3 = 8$
$-x^2 + \frac{8}{27}x^3 - 8 = 0$
$8x^3 - 27x^2 - 216 = 0$
Using a graphing calculator to solve this equation:
$x \approx 4.6334$
$x = 4.63 \text{ (to 3 s.f.)}$

1. Find the $x$-coordinates of the points on the curve where $y=2$.

Substitute $y=2$ into the curve equation:
$x^2 - 3x(2) + (2)^3 = 8$
$x^2 - 6x + 8 = 8$
$x^2 - 6x = 0$
$x(x - 6) = 0$
$x = 0$ or $x = 6$
The points are $(0, 2)$ and $(6, 2)$.

2. Differentiate the curve equation implicitly to find an expression for $\frac{dy}{dx}$.

Differentiating implicitly with respect to $x$ (as derived in part i):
$2x - \left( 3y + 3x\frac{dy}{dx} \right) + 3y^2\frac{dy}{dx} = 0$
$2x - 3y - 3x\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0$
$\frac{dy}{dx}(3y^2 - 3x) = 3y - 2x$
$\frac{dy}{dx} = \frac{3y - 2x}{3y^2 - 3x}$

3. Find the gradient of the normal and the equation of the normal at $(0, 2)$.

At $(0, 2)$:
$m_{\text{tangent}} = \frac{dy}{dx} = \frac{3(2) - 2(0)}{3(2)^2 - 3(0)} = \frac{6}{12} = \frac{1}{2}$
The gradient of the normal is $-\frac{1}{m_{\text{tangent}}} = -2$.
The equation of the normal is:
$y - 2 = -2(x - 0)$
$y = -2x + 2$

4. Find the gradient of the normal and the equation of the normal at $(6, 2)$.

At $(6, 2)$:
$m_{\text{tangent}} = \frac{dy}{dx} = \frac{3(2) - 2(6)}{3(2)^2 - 3(6)} = \frac{6 - 12}{12 - 18} = \frac{-6}{-6} = 1$
The gradient of the normal is $-\frac{1}{m_{\text{tangent}}} = -1$.
The equation of the normal is:
$y - 2 = -1(x - 6)$
$y - 2 = -x + 6$
$y = -x + 8$