No f(x)=x/3 -2/3

Let $y = f(x)$.

$$y = \frac{x}{3} - \frac{2}{3}$$

Swap $x$ and $y$:

$$x = \frac{y}{3} - \frac{2}{3}$$

Add $\frac{2}{3}$ to both sides:

$$x + \frac{2}{3} = \frac{y}{3}$$

Multiply by 3:

$$3\left(x + \frac{2}{3}\right) = y$$ $$3x + 2 = y$$

So, $f^{-1}(x) = 3x + 2$.

$$f(x) \cdot f^{-1}(x) = \left(\frac{x}{3} - \frac{2}{3}\right) \cdot (3x + 2)$$

We can factor out $\frac{1}{3}$ from the first term:

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

Now, multiply the binomials $(x-2)$ and $(3x+2)$:

$$f(x) \cdot f^{-1}(x) = \frac{1}{3}(x \cdot 3x + x \cdot 2 - 2 \cdot 3x - 2 \cdot 2)$$ $$f(x) \cdot f^{-1}(x) = \frac{1}{3}(3x^2 + 2x - 6x - 4)$$ $$f(x) \cdot f^{-1}(x) = \frac{1}{3}(3x^2 - 4x - 4)$$

Distribute the $\frac{1}{3}$:

$$f(x) \cdot f^{-1}(x) = \frac{3x^2}{3} - \frac{4x}{3} - \frac{4}{3}$$ $$f(x) \cdot f^{-1}(x) = x^2 - \frac{4x}{3} - \frac{4}{3}$$

The product of $f(x)$ and its inverse is:

$$x^2 - \frac{4x}{3} - \frac{4}{3}$$