To find $$(g \circ g \circ g)(4),$$ we first need to understand the composition of $$f$$ with itself, i.e., $$(f \circ f)(x) = f(f(x)) = g(x).$$ We can then repeatedly apply $$g$$ to get the given expression.

First, let's calculate $$(f \circ f)(x) = g(x):$$

$$g(x) = (f \circ f)(x) = f(f(x))$$

$$= f\left(\frac{4x+3}{6x-4}\right)$$

To evaluate this expression, we substitute $$\frac{4x+3}{6x-4}$$ for $$x$$ in the function $$f(x):$$

$$g(x) = f\left(\frac{4x+3}{6x-4}\right) = \frac{4\left(\frac{4x+3}{6x-4}\right) + 3}{6\left(\frac{4x+3}{6x-4}\right) - 4}$$

Now, we simplify the expression:

$$g(x) = \frac{4(4x+3) + 3(6x-4)}{6(4x+3) - 4(6x-4)}$$

$$= \frac{16x + 12 + 18x - 12}{24x + 18 - 24x + 16}$$

$$= \frac{34x}{34}$$

$$= x$$

So, $$g(x) = x$$ for all $$x$$ in the domain of $$g$$, which is $$\mathbb{R}-\left\{\frac{2}{3}\right\}$$. It's important to note that the domain restriction is preserved through the composition because $$f(x)$$ has a vertical asymptote at $$x = \frac{2}{3}$$ which doesn't intersect the graph.

So, $$g(x)$$ is the identity function on its domain, which means that applying $$g$$ any number of times will result in the same input for $$x$$ in the given domain. Hence, we have:

$$ (g \circ g \circ g \circ g)(4) = g(g(g(g(4)))) = g(g(g(4))) = g(g(4)) = g(4) = 4 $$

This corresponds to option D, which is $$4$$.