A question related to the normality of a ring 
Consider the ring $R=k[x^2 , xy,y^2 ]\subset k[x,y]$. My goal is to show that $R$ is normal but not a UFD.

Here is what I have gathered so far:
$(1)$ One can show that $R\cong k[x,y,z]/(x^2 -yz)$. Since $ k[x,y,z]/(x^2 -yz)$ is not a UFD, it follows then that $R$ is not a UFD as well.
$(2)$ I am still trying to show that $R$ is normal. It is easy to see that since $k[x,y,z]$ is a UFD and $xy-z^2$ is irreducible in $k[x,y,z]$, it follows that $(x^2-yz)$ is a prime ideal, hence $k[x,y,z]/(x^2 -yz)$ is an integral domain. That is $R$ is an integral domain. In order to show that $R$ is normal, I need to show that $R$ is its own integral closure in its field of fractions. But I am not sure how to prove this claim.
Any hint/help will be appreciated. Thanks in advance.
 A: If $\operatorname{char}k\ne2$ the normality follows from a general result; see Matsumura, Commutative Ring Theory, Example 4, page 65.
In general, one can show that $$k[x^2,xy,y^2]=k[x,y]\cap k(x^2,xy,y^2).$$ In order to do this notice that  $k[x^2,xy,y^2]$ is the subring of $k[x,y]$ consisting of polynomials which are sums of homogeneous polynomials of even degree (or, if you like, all the monomials which occur in $f$ have even degree). If $f\in k[x,y]\cap k(x^2,xy,y^2)$ then we can write $f(x,y)=g(x^2,xy,y^2)/h(x^2,xy,y^2)$. It follows that $f(x,y)h(x^2,xy,y^2)=g(x^2,xy,y^2)$. Now write these polynomials as sums of homogeneous polynomials in $x$ and $y$: $f=f_0+f_1+\dots+f_m$, $g=g_0+g_2+\dots+g_{2r}$, and $h=h_0+h_2+\dots+h_{2s}$. From $fh=g$ we get $f_mh_{2s}=g_{2r}$, which shows that $m$ is even, that is, $f_m\in k[x^2,xy,y^2]$. Now notice that $f-f_m\in k[x,y]\cap k(x^2,xy,y^2)$, and  recurrently we get that $f$ is a sum of homogeneous polynomials of even degree, hence $f\in k[x^2,xy,y^2]$.
$R=k[x^2,xy,y^2]$ is not a UFD since $x^2y^2=(xy)^2$ and $x^2,xy,y^2$ are irreducible in $R$ (why?).
