Yours is a very interesting and subtle question, which often generates confusion. First let us give a name to the property you are interested in: a ring $A$ will be said to satisfy (DIM) if for all $\mathfrak p \in \operatorname{Spec}(A)$ we have $$\operatorname{height}(\mathfrak p) +\dim A/\mathfrak p=\dim(A) \quad \quad (\text{DIM})$$
The main misconception is to believe that this follows from catenarity:
Fact 1: A catenary ring, or even a universally catenary ring, does not satisfy (DIM) in general.
Counterexample: Let $(R,\mathfrak m)$ be a discrete valuation ring whose maximal ideal has uniformizing parameter $\pi$, i.e. $\mathfrak m =(\pi)$. Let $A=R[T]$, the polynomial ring over $R$. The ring $A$ has dimension $2.$ Then for the maximal ideal $\mathfrak p=(\pi T-1)$, the relation (DIM) is false:
$\operatorname{height}(\mathfrak p)+\dim A/\mathfrak p= 1+0=1\neq 2=\dim (A)$.
And this even though $A$ is as nice as can be: an integral domain, noetherian, regular, universally catenary,...
Happily here are two positive results:
Fact 2: A finitely generated integral algebra over a field satisfies (DIM) (and is universally catenary).
So, by the algebro-geometric dictionary, an affine variety $X$ has the pleasant property that for each integral subvariety $Y\subset X$ we have, as hoped, $\operatorname{dimension}(Y) + \operatorname{codimension}(Y)$ $=$ $\operatorname{dimension}(X).$
Fact 3: A Cohen-Macaulay local ring satisfies (DIM) (and is universally catenary).
For example a regular ring is Cohen-Macaulay. This "explains" why my counter-example above was not local.
The paradox resolved. How is it possible for a catenary ring $A$ not to satisfy (DIM)? Here is how. If you have an inclusion of two primes $\mathfrak p\subsetneq \mathfrak q$ catenarity says that you can complete it to a saturated chain of primes
$\mathfrak p\subsetneq \mathfrak p_1\subsetneq \ldots \subsetneq \mathfrak p_{r-1} \subsetneq \mathfrak q$ and that all such completions will have length the same length $r$. Fine. But what can you say if you have just one prime $\mathfrak p$ ? Not much! The catenary ring $A$ may have dimension $\dim(A) > \operatorname{height}( \mathfrak p) +\dim(A/\mathfrak p)$ because it possesses a long chain of primes avoiding the prime $\mathfrak p$ altogether. In my counterexample above the only saturated chain of primes containing $\mathfrak p=(\pi T-1)$ is $0\subsetneq \mathfrak p$. However the ring $A$ has dimension 2 because of the saturated chain of primes $0\subsetneq (\pi) \subsetneq (\pi,T)$, which avoids $\mathfrak p$.
Addendum. Here is why the ideal $\mathfrak p$ in the counter-example is maximal. We have $A/\mathfrak p=R[T]/(\pi T-1)=R[1/\pi]=\operatorname{Frac}(R)$, since the fraction field of a discrete valuation ring can be obtained just by inverting a uniformizing parameter. So $A/\mathfrak p$ is a field and $\mathfrak p$ is maximal.
This is true - in a Noetherian ring, the only possible heights for which there are finitely many primes of that height are either $0$, or maximal ideals. This follows from the following fact:
Proposition: If $R$ is Noetherian and $\mathfrak p_0 \subsetneq \mathfrak p_1 \subsetneq \mathfrak p_2$ is a chain of distinct prime ideals in $R$, then there are infinitely many primes $\mathfrak q$ such that $\mathfrak p_0 \subsetneq \mathfrak q \subsetneq \mathfrak p_2$.
Proof: Localizing at $\mathfrak p_2$ and quotienting by $\mathfrak p_0$, it suffices to show that any Noetherian local domain of dimension $2$ has infinitely many primes. If there were only finitely many height $1$ primes, then since $\mathfrak p_2$ is not contained in any of them, by prime avoidance it is not contained in their union, so there exists $x \in \mathfrak p_2$, but not in any height $1$ prime. But this contradicts the Principal Ideal Theorem, since $\text{ht}(x) \le 1$.
Best Answer
Since $\mathbf{C}[\gamma^2,\gamma^3]\subset \mathbf{C}[\gamma]$ is an integral extension we obtain $\dim\mathbf{C}[\gamma^2,\gamma^3]=\dim\mathbf{C}[\gamma]=?$.
$\mathbf{C}[\alpha,\beta,\gamma]/(\gamma-\alpha\beta)\simeq\mathbf{C}[\alpha,\beta]$ while $\mathbf{C}[\alpha,\beta]/(\alpha^2+\beta,\alpha^3\beta^2)\simeq \mathbf{C}[\alpha]/(\alpha^7)$, so the first dimension is ... and the second is ....