If $$y(x) = x^x, x > 0$$, then $$y''(2) - 2y'(2)$$ is equal to:

Given $$y(x) = x^x$$, $$x > 0$$. We need to find $$y''(2) - 2y'(2)$$.

Taking the natural logarithm of both sides:

$$\ln y = x \ln x$$

Differentiating both sides with respect to $$x$$:

$$\frac{y'}{y} = \ln x + 1$$

$$y' = x^x(\ln x + 1)$$

Differentiating again:

$$y'' = \frac{d}{dx}[x^x(\ln x + 1)]$$

$$= x^x(\ln x + 1)^2 + x^x \cdot \frac{1}{x}$$

$$= x^x\left[(\ln x + 1)^2 + \frac{1}{x}\right]$$

Evaluating at $$x = 2$$:

$$y(2) = 2^2 = 4$$

$$y'(2) = 4(\ln 2 + 1)$$

$$y''(2) = 4\left[(\ln 2 + 1)^2 + \frac{1}{2}\right]$$

Computing $$y''(2) - 2y'(2)$$:

$$= 4\left[(\ln 2 + 1)^2 + \frac{1}{2}\right] - 2 \cdot 4(\ln 2 + 1)$$

$$= 4\left[(\ln 2)^2 + 2\ln 2 + 1 + \frac{1}{2} - 2\ln 2 - 2\right]$$

$$= 4\left[(\ln 2)^2 - \frac{1}{2}\right]$$

$$= 4(\log_e 2)^2 - 2$$

The answer is Option C: $$4(\log_e 2)^2 - 2$$.