Question about regularity / integrality 
Let $k$ be an algebraically closed  field, and let $R:=k[X,Y,Z]/(X^2+Y^5+Z^3)$. Show that, for any $f\in (X,Y)\subset k[X,Y]$, the localization $R[f^{-1}]$ is integrally closed,  and deduce  that $R$ is  integrally  closed.

First, observe that $X^2+Y^5+Z^3$ is irreducible, for in $k[X,Y]$ there are no cubic roots of  $X^2+Y^5$, and so it is the minimal polynomial of $\sqrt[3]{X^2+Y^5}$ over $k[X,Y]$. This means that $R$ is a domain and a free $k[X,Y]$-module. We will need also that $\operatorname{Spec}R$ is regular at every  point but $(0,0,0)$, that is easily verified with the Jacobian criterion, knowing $\operatorname{dim}R=2$. Thus a localization $R_\mathfrak M$, for any maximal ideal $\mathfrak M\subset R$, is regular if and only $\mathfrak M$ is not (the image in $R$ of) $(X,Y,Z)$.
Claim: if $S$ is a multiplicatively closed set of a domain $R$, $\mathfrak  p\subset R[S^{-1}]$ is a prime ideal, and $\mathfrak P:=\mathfrak p\cap  R$, then  $R[S^{-1}]_{\mathfrak p}\cong R_{\mathfrak P}$. Proof: for  every subset $A\subseteq R$, denote with $A[S^{-1}]$ the set of (the equivalence classes of) the elements in $R[S^{-1}]$ with numerators in $A$; set also $T:=R-\mathfrak  P$,  so  that $R_{\mathfrak P}\cong R[T^{-1}]$.  We know that $\mathfrak p=\mathfrak P[S^{-1}]$ and $R[S^{-1}]-\mathfrak P[S^{-1}]=T[S^{-1}]$, so $R[S^{-1}]_{\mathfrak p}\cong R[S^{-1}][T[S^{-1}]]^{-1}$. Since $S\subseteq T$, it makes sense to define this map, that is actually a isomorphism: $$R[S^{-1}][T[S^{-1}]]^{-1}\to R[T^{-1}]:\frac rs /\frac t{s'}\mapsto  \frac{rs'}{st}.$$
Then we can say that  for any maximal ideal $\mathfrak m\subset R[f^{-1}]$ there is an isomorphism $R[f^{-1}]_\mathfrak m\cong R_\mathfrak M$, where $\mathfrak M:=\mathfrak m\cap R$. But $R_\mathfrak M$ is regular unless $\mathfrak M=(X,Y,Z)$, and this cannot be, as $f\notin \mathfrak M$, so $R[f^{-1}]_\mathfrak m$ is regular for any $\mathfrak m$. Here I have a question: does regularity (of a domain) implies integral  closedness? If yes, we'd be done since the localization of $R[f^{-1}]$ at any maximal ideal  would be integrally closed; however in my course we only saw that the two conditions are equivalent in dimension $1$, and I didn't find results in this direction on the books of Atiyah&Macdonald or Bosch.
In order to see that $R$ is integrally closed,  we can show that $R=\bigcap_{f\in (X,Y)}R[f^{-1}]$; then we'd be done, for the intersection of integrally closed domains is integrally closed. Notice that $k[X,Y]_f=\{\frac hg\in k(X,Y):g=f^n\text{ for some }n\}$, meaning that $\bigcap_{f\in (X,Y)}k[X,Y]_f$ is the set $\{\frac hg\in k(X,Y):\forall f\in (X,Y)\ (g=f^n\text{ for some }n)\}$; however using that $k[X,Y]$ is a factorial ring it should  easily follow that the only element that is power of every $f\in (X,Y)$ is $1$. Hence $\bigcap_{f\in (X,Y)}k[X,Y]_f=k[X,Y]$; knowing that $R\cong k[X,Y]\oplus k[X,Y]\oplus k[X,Y]$, finally $$\bigcap_{f\in(X,Y)}R[f^{-1}]\cong \bigcap_{f\in(X,Y)}k[X,Y]_f\oplus k[X,Y]_f\oplus k[X,Y]_f\cong k[X,Y]\oplus k[X,Y]\oplus k[X,Y]\cong R.$$
So the main question is the one in italics; anyway you're free to read the (attempt of) proof, any correction is welcome.
 A: Yes, regularity implies integral closedness. By the Auslander-Buchsbaum theorem, a regular local ring is a UFD, and it is well-known that a UFD is integrally closed in its field of fractions. So the localization of the domain $R[f^{-1}]$ at any maximal ideal is integrally closed in its field of fractions, which implies that $R[f^{-1}]$ must be integrally closed in its field of fractions too. (More generally, a regular scheme is normal, and for any open $U\subset X$ in a normal scheme, $\mathcal{O}_X(U)$ is a normal ring, ref 033J.)
One may also conclude that $R$ is integrally closed from the Serre conditions $R_1$ and $S_2$ - $S_2$ is automatic from being a hypersurface in affine space, and $R_1$ can be shown by computing the singular locus via the Jacobian criteria. This is also well-covered in the literature and on MSE.
Let me also mention that this ring displays some interesting behavior: if $k$ is not of characteristic 2, 3, or 5, the completion of this ring, $k[[x,y,z]]/(x^2+y^3+z^5)$, is the only nonregular normal complete 2-dimensional local ring which is a UFD. This was proven by Lipman in Rational singularities with applications to algebraic surfaces and unique factorization. The spectrum of $R$ is also a Du Val singularity, and the Dynkin diagram you get by resolving it is $E_8$, which is another interesting property.
