Since x=3 is a vertical asymptote, then x−3 is a factor in the denominator. Since there is a removable discontinuity at x=−2, then x+2 is a common factor in the numerator and denominator. So, a possible rational function is:

\(\displaystyle{f{{\left({x}\right)}}}=\frac{{{x}+{2}}}{{{\left({x}-{3}\right)}{\left({x}+{2}\right)}}}\)

or

\(\displaystyle{f{{\left({x}\right)}}}=\frac{{{x}+{2}}}{{{x}^{{2}}-{x}-{6}}}\)

\(\displaystyle{f{{\left({x}\right)}}}=\frac{{{x}+{2}}}{{{\left({x}-{3}\right)}{\left({x}+{2}\right)}}}\)

or

\(\displaystyle{f{{\left({x}\right)}}}=\frac{{{x}+{2}}}{{{x}^{{2}}-{x}-{6}}}\)