The correct answer is $\boxed{\frac{1}{2},\frac{1}{3}}$.
The poles of a discrete-time system are the roots of the denominator of the system transfer function. In this case, the system transfer function is given by
$$H(z) = \frac{1}{1 – \frac{1}{2}z^{-1} – \frac{1}{3}z^{-2}}$$
The poles of the system are the roots of the denominator polynomial, which is given by
$$1 – \frac{1}{2}z^{-1} – \frac{1}{3}z^{-2} = (z – \frac{1}{2})(z – \frac{1}{3})$$
Therefore, the poles of the system are located at $\frac{1}{2}$ and $\frac{1}{3}$.
Option A is incorrect because the poles are not located at $2$ and $3$. Option B is incorrect because the poles are not located at $\frac{1}{2}$ and $3$. Option C is incorrect because the poles are not located at $\frac{1}{2}$ and $\frac{1}{3}$.