This question is very related to this other question. I have an alternative solution to the ones proposed in the answers, and I'd like to know if it is correct.

I want to find the dimension of $A=\mathbb{C}[x,y]/(x^3-y^2)$ using the dimension theory for Noetherian local rings that is extensively developed in chapter $11$ of the book *Introduction to Commutative Algebra* by Atiyah and MacDonald. If $R$ is any Noetherian ring we have:
$$\dim R = \sup_{m\subset R\mathrm{\ max.\ id.}}\dim R_m$$
(since every chain of primes is contained in a maximal ideal). Let $m\subset A$ be a maximal ideal. If $x^3-y^2\notin m$, then $A_m=0$ and thus $\dim A_m=0$, else if $x^3-y^2\in m$ we use exactness of the localization functor to see that:
$$A_m=\mathbb{C}[x,y]_m/(x^3-y^2)_m=\mathbb{C}[x,y]_m/(x^3-y^2)$$
where we denote again by $x^3-y^2$ the element $\frac{x^3-y^2}{1}\in\mathbb{C}[x,y]_m$. Since $x^3-y^2$ is not a zero-divisor in $\mathbb{C}[x,y]_m$, we have (by corollary $11.18$ in Atiyah/MacDonald):
$$\dim A_m=\dim \mathbb{C}[x,y]_m - 1=1$$
Thus $\dim A=1$.

Is this a valid way to proceed or am I doing something illicit? Also, would this work also for quotients by ideals generated by more than one element? For example, say I'd like to compute $\dim \mathbb{C}[x,y,z]/(x^2+y,x^3y^2)$. I would use the fact that: $$A/(f,g)\cong[A/(f)]/[(f,g)/(f)]$$ Then I would compute the dimension of $A/(f)$ as above. $(f,g)/(f)=(\bar{g})$ (generated by the image of $g$ in $A/(f)$), and repeat.