Step 1: Differentiate the equation of the curve $6y = x^3 + 2$ with respect to time $t$.

$$6\frac{dy}{dt} = 3x^2 \frac{dx}{dt}$$

Step 2: Apply the given condition. The problem states that the $y$-coordinate changes 8 times as fast as the $x$-coordinate, which means $\frac{dy}{dt} = 8 \frac{dx}{dt}$. Substitute this into the differentiated equation:

$$6\left(8 \frac{dx}{dt}\right) = 3x^2 \frac{dx}{dt}$$

Step 3: Simplify the equation to solve for $x$. Assuming $\frac{dx}{dt} \neq 0$, we can divide both sides by $\frac{dx}{dt}$:

$$48 = 3x^2$$ $$x^2 = 16$$ $$x = \pm 4$$

Step 4: Find the corresponding $y$-coordinates by substituting the values of $x$ into the original curve equation $6y = x^3 + 2$. For $x = 4$:

$$6y = (4)^3 + 2 = 64 + 2 = 66$$ $$y = 11$$

Point is $(4, 11)$.

For $x = -4$:

$$6y = (-4)^3 + 2 = -64 + 2 = -62$$ $$y = \frac{-31}{3}$$

Point is $\left(-4, \frac{-31}{3}\right)$.