= Solution
An <integral extension> $R\subseteq T$ means that every element $t\in T$ satisfies a <monic polynomial> with coefficients in $R$:
$$
t^n+r_{n-1}t^{n-1}+\cdots+r_0=0.
$$
The <Krull dimension> is the supremum of the lengths of strict chains of primes:
$$
\boxed{\dim R=\sup\{n:\mathfrak p_0\subsetneq\mathfrak p_1\subsetneq\cdots\subsetneq\mathfrak p_n\}.}
$$
There need not be a finite bound on these lengths.
Here are the prime-ideal facts behind dimension preservation, with their relevant proofs. If a domain $B$ is integral over a subdomain $A$ and $B$ is a <field>, then $A$ is a <field>: for $0\neq a\in A$, a monic equation for $a^{-1}\in B$, multiplied by a suitable power of $a$, expresses $a^{-1}$ as an element of $A$. Conversely, an <integral domain> integral over a <field> is itself a <field>: a nonzero element has a polynomial equation with nonzero constant term after removing any factor of the indeterminate, and that equation expresses its inverse. Applied to quotients, these observations show that a prime in an <integral extension> is maximal if and only if its contraction is maximal.
For the <Lying-over theorem>, localize $R\subseteq T$ at $S=R\setminus\mathfrak p$. The inclusion remains injective and integral, and $S^{-1}T$ is nonzero. Any <maximal ideal> of $S^{-1}T$ contracts to the unique <maximal ideal> of $R_{\mathfrak p}$, by the <field> criterion just proved. The <prime ideal correspondence for localization> then gives a prime of $T$ contracting to $\mathfrak p$.
For the <Going-up theorem>, suppose $\mathfrak q$ lies over $\mathfrak p$ and $\mathfrak p\subseteq\mathfrak p'$. The quotient inclusion $R/\mathfrak p\subseteq T/\mathfrak q$ is integral. Apply <Lying-over theorem> to the prime $\mathfrak p'/\mathfrak p$; lifting back gives $\mathfrak q'\supseteq\mathfrak q$ contracting to $\mathfrak p'$.
For the <incomparability theorem for integral extensions>, suppose $\mathfrak q\subseteq\mathfrak q'$ contract to the same $\mathfrak p$. After localizing at $R\setminus\mathfrak p$, both are maximal <ideals>, because they lie over the <maximal ideal> of $R_{\mathfrak p}$. Their inclusion is therefore equality. The bijection between primes under localization gives $\mathfrak q=\mathfrak q'$.
Now contract a strict chain of primes in $T$. <Incomparability theorem for integral extensions> ensures that every contraction remains strict, so $\dim T\leq\dim R$. Conversely, <Lying-over theorem> lifts the first member of any finite chain in $R$, and repeated <Going-up theorem> lifts the remaining members; different contractions ensure a strict chain in $T$. Taking suprema, including the possibility of infinity, gives
$$
\boxed{\dim T=\dim R.}
$$
This proves that <integral extensions preserve Krull dimension>.
For the given quotient, put $B=k[Y][X]/(X^2+YX+Y^3)$. The relation is monic in $X$, so <monic polynomial> division gives a unique representative $a(Y)+b(Y)X$. Thus $k[Y]\hookrightarrow B$ is injective and $B$ is free of rank two as a $k[Y]$-module. In particular, $B$ is integral over $k[Y]$. The one-variable polynomial <ring> has dimension one: its zero prime is strictly below $(Y)$, and every nonzero prime is maximal because $k[Y]$ is a principal <ideal> domain. Hence
$$
\boxed{\dim k[X,Y]/(XY+X^2+Y^3)=1.}
$$
This is an instance of <dimension of a monic plane hypersurface>. No irreducibility or algebraic-closure assumption on $k$ is needed; monicity supplies the <integral extension> in every characteristic.
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