1. We are given $f(x)=\frac{4x+3}{6x-4}, \qquad x\ne \frac{2}{3}$ and $g(x)=(f\circ f)(x)=f(f(x)).$

   We need to find $(g\circ g\circ g)(4).$

2. First compute $f(f(x))$.

   Let $y=f(x)=\frac{4x+3}{6x-4}.$ Then $f(f(x))=f(y)=\frac{4y+3}{6y-4}.$

   Substitute $y=\dfrac{4x+3}{6x-4}$: $f(f(x))=\frac{4\left(\frac{4x+3}{6x-4}\right)+3}{6\left(\frac{4x+3}{6x-4}\right)-4}.$

3. Simplify the numerator:

   =\frac{16x+12}{6x-4}+3 =\frac{16x+12+3(6x-4)}{6x-4} =\frac{16x+12+18x-12}{6x-4} =\frac{34x}{6x-4}.$$

4. Simplify the denominator:

   =\frac{24x+18}{6x-4}-4 =\frac{24x+18-4(6x-4)}{6x-4} =\frac{24x+18-24x+16}{6x-4} =\frac{34}{6x-4}.$$

5. Therefore, $f(f(x))=\frac{\frac{34x}{6x-4}}{\frac{34}{6x-4}}=x.$

   Hence, $g(x)=x.$ So $g$ is the identity function on its domain.

6. Now compute repeated composition: $g(g(g(4)))=4.$

7. Check options:

   • A: $-4$ ❌
   • B: $\frac{19}{20}$ ❌
   • C: $-\frac{19}{20}$ ❌
   • D: $4$ ✅

Therefore, the correct answer is $\boxed{4}.$