The problem states that the maximum speed of a car on a curved road is inversely proportional to the square root of the curvature. The curvature is given by the equation $y = \frac{1}{3}x^3$, and the speed for a certain point on the curve $(1, \frac{1}{3})$ is given as 30 miles per hour. The task is to find the speed at another point $\left(\frac{9}{5}, \frac{243}{125}\right)$.

Step-by-step solution:

1. Curvature and Proportionality: The speed $v$ is inversely proportional to the square root of the curvature $\kappa$: $v \propto \frac{1}{\sqrt{\kappa}}$ This can be written as: $v = \frac{k}{\sqrt{\kappa}}$ where $k$ is some constant of proportionality.

2. Curvature of the curve $y = \frac{1}{3}x^3$: The curvature $\kappa$ of a function $y = f(x)$ is given by the formula: $\kappa = \frac{|f''(x)|}{(1 + (f'(x))^2)^{3/2}}$

   First, calculate the first and second derivatives of $y = \frac{1}{3}x^3$: $f'(x) = x^2$ $f''(x) = 2x$

3. Curvature at $x = 1$: $\kappa_1 = \frac{|2(1)|}{(1 + (1)^4)^{3/2}} = \frac{2}{(1+1)^{3/2}} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$

   Since the car's speed is given as 30 mi/h at $x = 1$, we can find $k$: $30 = \frac{k}{\sqrt{\frac{1}{\sqrt{2}}}} \Rightarrow k = 30\sqrt{\frac{1}{\sqrt{2}}}$

4. Find curvature at $x = \frac{9}{5}$:

   Substituting $x = \frac{9}{5}$ into the curvature formula: $f'(x) = \left(\frac{9}{5}\right)^2 = \frac{81}{25}$ and $f''(x) = 2 \times \frac{9}{5} = \frac{18}{5}$ Therefore: $\kappa_2 = \frac{\frac{18}{5}}{\left(1 + \left(\frac{81}{25}\right)\right)^{3/2}} = \frac{\frac{18}{5}}{\left(1 + \frac{81}{25}\right)^{3/2}}$
   Simplify: [ \kappa_2 = \frac{\frac{18}{5}}{\left(\frac{106}{25}\right)^{3/2}} = \frac{\frac{18}{5}}{\left(\frac{106^{3/2}}{25^{3/2}}\right)}