A particle moves along a curve $y=\frac{2 x^3-1}{3}$. The points on the curve at which the $y$ co-ordinate is changing 18 times the $x$ co-ordinate are

The curve is described by $y = \frac{2x^3 - 1}{3}$, with the condition that $\frac{dy}{dt} = 18\frac{dx}{dt}$.

Differentiating implicitly with respect to $t$ gives $\frac{dy}{dt} = 2x^2\frac{dx}{dt}$.

Equating the two expressions for $\frac{dy}{dt}$ yields $18\frac{dx}{dt} = 2x^2\frac{dx}{dt}$. Since $\frac{dx}{dt} \neq 0$, this simplifies to $18 = 2x^2$ or $x^2 = 9$, giving $x = \pm 3$.

Substituting into the original equation: at $x = 3$, $y = \frac{2(27) - 1}{3} = \frac{53}{3}$; at $x = -3$, $y = \frac{2(-27) - 1}{3} = -\frac{55}{3}$.

The corresponding points are $\left(3, \frac{53}{3}\right)$ and $\left(-3, -\frac{55}{3}\right)$, which correspond to option D.