To find the domain of \( f(x)=\dfrac{\sqrt{x^2-3x-4}}{6x^2-54}\) we know \(x^2-3x-4\geq 0\) and \(6x^2-54 \neq 0\). \(x^2-3x-4\geq 0\) for \((-\infty, -1]\cup[4,\infty)\) and \(6x^2-54\neq 0\) for \((-\infty,-3)\cup(-3,3)\cup(3,\infty)\). So the domain of \(f(x)\) is \( (-\infty,-3)\cup (-3,-1]\cup [4,\infty).\)