A function will be invertible (on a proper domain) if and only if it is injective, that is, if it never takes the same value twice. If your function is continuous and is defined on an interval, this forces it to be increasing or decreasing. Even without being defined on an interval we can use calculus to determine if the function is increasing/decreasing, which lets us solve the problem fairly simply.
We have that f'(x)=\frac{(x-3)-(x+2)}{(x-3)^2}=\frac{-5}{(x-3)^2}. This is negative except where it is undefined, and so the function is decreasing, at least locally. Unfortunately, because we have the discontinuity at $x=3$, we can't say immediately that the function is injective. However, we do know that it never takes the same value twice on either side of the discontinuity. This combined with limits, is all we need:
Since $\displaystyle \lim_{x\to -\infty}f(x)=\lim_{x\to \infty}f(x)=1$, we must have that $f(x)<1$ if $x<3$ and $f(x)>1$ if $x>3$. Since no number is both less than 1 and greater than 1, we can't have $f(x_1)=f(x_2)$ for $x_1<3$ and $x_2>3$. Therefore $f(x)$ is injective, and hence invertible on a suitable domain (namely on $(-\infty,3)\cup (3,\infty)$.)