It is given that $f'(x)=6\cos 3x-12\sin 3x$ and $f\left(\frac{\pi}{3}\right)=0$.

1. Integrate $f'(x)$ to find the general form of $f(x)$.

$f(x) = \int (6\cos 3x - 12\sin 3x) \, dx$
$f(x) = 6\left(\frac{\sin 3x}{3}\right) - 12\left(-\frac{\cos 3x}{3}\right) + c$
$f(x) = 2\sin 3x + 4\cos 3x + c$

2. Use the given condition $f\left(\frac{\pi}{3}\right) = 0$ to find the constant of integration $c$.

$f\left(\frac{\pi}{3}\right) = 2\sin\left(3 \cdot \frac{\pi}{3}\right) + 4\cos\left(3 \cdot \frac{\pi}{3}\right) + c = 0$
$2\sin(\pi) + 4\cos(\pi) + c = 0$
$2(0) + 4(-1) + c = 0$
$-4 + c = 0$
$c = 4$

3. State the final function $f(x)$.

$f(x) = 2\sin 3x + 4\cos 3x + 4$

1. Differentiate $f'(x)$ to find the second derivative $f''(x)$.

$f''(x) = \frac{d}{dx}(6\cos 3x - 12\sin 3x)$
$f''(x) = 6(-3\sin 3x) - 12(3\cos 3x)$
$f''(x) = -18\sin 3x - 36\cos 3x$

2. Substitute $x = \frac{\pi}{12}$ and evaluate using exact trigonometric values.

$f''\left(\frac{\pi}{12}\right) = -18\sin\left(3 \times \frac{\pi}{12}\right) - 36\cos\left(3 \times \frac{\pi}{12}\right)$
$= -18\sin\left(\frac{\pi}{4}\right) - 36\cos\left(\frac{\pi}{4}\right)$
$= -18\left(\frac{\sqrt{2}}{2}\right) - 36\left(\frac{\sqrt{2}}{2}\right)$
$= -9\sqrt{2} - 18\sqrt{2}$
$= -27\sqrt{2}$