Answer
$(-\infty,-2)\cup(-2,3)\cup(3,\infty)$.
Work Step by Step
The domain requirement for $f(x)=log(\frac{x+2}{x-3})^2$ is $x-3\ne0, x+2\ne0$ which gives $x\ne-2,3$ or in interval notation $(-\infty,-2)\cup(-2,3)\cup(3,\infty)$.
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