I'm looking for verification of whether the following solution is correct.
Let $R=\mathbb{Q}[x,y,z]/(xy-z^3),\ I=(x,y,z)$ and $A=R_I$. Since localization commutes with taking quotients, $A=\mathbb{Q}[x,y,z]_I / (xy-z^3)$. The ring $\mathbb{Q}[x,y,z]_I$ is local and Noetherian, and $xy-z^3$ is not a zero-divisor, so Corollary 10.18 of Atiyah-MacDonald implies $\dim A=\dim \mathbb{Q}[x,y,z]_I-1$.
Since $\dim \mathbb{Q}[x,y,z]=3$, $\dim \mathbb{Q}[x,y,z]_I \leq 3$. The chain $(0) \subset (x) \subset (x,y) \subset (x,y,z)$ of prime ideals of $\mathbb{Q}[x,y,z]_I$ shows that the dimension is $3$. We know these are all prime ideals in the localisation since they are prime ideals in $\mathbb{Q}[x,y,z]$ contained in $(x,y,z)$. Therefore $\dim A= 2.$