Given the function:

$$ f(x) = x - 2\sin{x}\cos{x} + \frac{1}{3}\sin{3x} $$

We want to find the maximum value of this expression for $$0 \leq x \leq \pi$$.

Step 1: Rewrite the expression

Notice that we can rewrite the expression as:

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

Step 2: Find the first derivative

Now, let's find the derivative of this expression with respect to $$x$$:

$$ f'(x) = 1 - 2\cos{2x} + \cos{3x} $$

Step 3: Find the critical points

To find the critical points, we set $$f'(x) = 0$$:

$$ 1 - 2\cos{2x} + \cos{3x} = 0 $$

Step 4: Solve for x

This equation can be rewritten as:

$$ (2\cos{x} + \sqrt{3})(2\cos{x} - \sqrt{3})(\cos{x} - 1) = 0 $$

We get three possible solutions for $$x$$:

$$ \cos{x} = \frac{-\sqrt{3}}{2}, \frac{\sqrt{3}}{2}, 1 $$

Which gives us: $$ x = \frac{5\pi}{6}, \frac{\pi}{6}, 0 $$

Step 5: Find the second derivative

Now, let's find the second derivative of the expression:

$$ f''(x) = 4\sin{2x} - 3\sin{3x} $$

Step 6: Analyze the critical points Evaluate the second derivative at the critical points:

1. $$f''\left(\frac{5\pi}{6}\right) = -2\sqrt{3} - \sqrt{3} < 0$$, indicating a local maximum.
2. $$f''\left(\frac{\pi}{6}\right) = 2\sqrt{3} - \sqrt{3} > 0$$, indicating a local minimum.
3. $$f''(0) = 0$$, inconclusive.

Step 7: Evaluate the function at the local maximum point and the boundary points Since we are looking for the maximum value of $$f(x)$$, we can now evaluate the function at the local maximum point and the boundary points:

1. $$f\left(\frac{5\pi}{6}\right) = \frac{5\pi}{6} + \frac{\sqrt{3}}{2} + \frac{1}{3} = \frac{5\pi + 2 + 3\sqrt{3}}{6}$$
2. $$f(0) = 0$$
3. $$f(\pi) = \pi$$

Step 8: Compare the values The maximum value of $$f(x)$$ is given by $$f\left(\frac{5\pi}{6}\right) = \frac{5\pi + 2 + 3\sqrt{3}}{6}$$.