#### Answer

$(-\infty,\ -3)\cup (-3,2)\cup (2,\ \infty)$

#### Work Step by Step

We are given:
$f(x)= \frac{x^{4}}{x^{2}+x-6}$
The domain of a function consists of all values that $x$ is allowed to have. In this case, we can not have division by $0$:
$x^{2}+x-6\neq 0$
$(x+3)(x-2)\neq 0$
$(x+3)\neq 0$ or $(x-2)\neq 0$
$ x\neq-3$ or $x\neq 2$
Thus the domain is: $(-\infty,\ -3)\cup (-3,2)\cup (2,\ \infty)$