The problem asks to analyze the given function $f(x) = \frac{2}{3}x^3 - x^2 - \frac{2}{3}$.

1. Problem Description

The problem asks to analyze the given function $f(x) = \frac{2}{3}x^3 - x^2 - \frac{2}{3}$.

2. Solution Steps

The function is $f(x) = \frac{2}{3}x^3 - x^2 - \frac{2}{3}$.

Since the problem asks us to solve it, without any other context, it is hard to provide a specific answer. I will find the derivative of this function.

First derivative:

$f'(x) = \frac{d}{dx}(\frac{2}{3}x^3 - x^2 - \frac{2}{3})$

$f'(x) = \frac{2}{3} \cdot 3x^2 - 2x - 0$

$f'(x) = 2x^2 - 2x$

Second derivative:

$f''(x) = \frac{d}{dx}(2x^2 - 2x)$

$f''(x) = 4x - 2$

To find critical points, we set the first derivative to zero:

$f'(x) = 2x^2 - 2x = 0$

$2x(x - 1) = 0$

$x = 0$ or $x = 1$

To determine if these are local maxima or minima, we use the second derivative test:

$f''(0) = 4(0) - 2 = -2 < 0$, so $x=0$ is a local maximum.

$f(0) = \frac{2}{3}(0)^3 - (0)^2 - \frac{2}{3} = -\frac{2}{3}$

$f''(1) = 4(1) - 2 = 2 > 0$, so $x=1$ is a local minimum.

$f(1) = \frac{2}{3}(1)^3 - (1)^2 - \frac{2}{3} = \frac{2}{3} - 1 - \frac{2}{3} = -1$

3. Final Answer

The derivative of the function is $f'(x) = 2x^2 - 2x$.

The second derivative is $f''(x) = 4x - 2$.

The critical points are at $x=0$ and $x=1$.

There is a local maximum at $(0, -\frac{2}{3})$ and a local minimum at $(1, -1)$.