To find the domain of the logarithmic function $f(x) = \log_{g(x)} h(x)$, we must satisfy three critical conditions:

1. The argument must be strictly positive: $h(x) > 0 \implies x^2 - 8x + 12 > 0$. Factoring this quadratic, we get $(x-2)(x-6) > 0$, which gives $x 6$.
2. The base must be strictly positive: $g(x) > 0 \implies x - 3 > 0 \implies x > 3$.
3. The base cannot be equal to $1$: $g(x) \neq 1 \implies x - 3 \neq 1 \implies x \neq 4$.

Now, we find the intersection of these three conditions:

• From (1), $x \in (-\infty, 2) \cup (6, \infty)$.
• From (2), $x \in (3, \infty)$.
• From (3), $x \neq 4$.

The intersection of $x \in (3, \infty)$ and $x \in (-\infty, 2) \cup (6, \infty)$ is $x \in (6, \infty)$. Since all values in $(6, \infty)$ are already greater than $4$, the condition $x \neq 4$ is automatically satisfied. Thus, the domain is $(6, \infty)$.