How to find ROC of system when input is two sided and output is one-sided
The transfer function is

$$H(z)=\frac{Y(z)}{X(z)}=-\frac18\frac{z-3}{\left(z-\frac14\right)(z-\frac{10}{3})}\tag{1}$$

It has three possible ROCs:


$|z|&lt;\frac14$ (left-sided, unstable)
$\frac14&lt;|z|&lt;\frac{10}{3}$ (two-sided, stable)
$|z|&gt;\frac{10}{3}$ (right-sided, unstable)


Only the second ROC, corresponding to a two-sided and stable impulse response, overlaps with the ROC of $X(z)$. The intersection, and, consequently, the ROC of $Y(z)$ would be $\frac13&lt;|z|&lt;3$ (equal to the ROC of $X(z)$). However, the pole at $z=3$ is removed by pole-zero cancellation, which results in the final ROC $|z|&gt;\frac13$ for $Y(z)$.