Show that the ring $k[x,y,z]/(x^2-yz)$ is normal 
Show that the ring $A=k[x,y,z]/(x^2-yz)$ is normal

My attempt:
Take an element $\frac{f}{g} \in K(A)=(k[x,y,z]/(x^2-yz))_{(x^2-yz)}$ assuming that it is integral over $A$. Since $z=\frac{x^2}y \in K(A)$ we can assume that $f$ and $g$ depend only on $x$ and $y$. There exist a monic polynomial $\left(\frac fg\right)^n+a_1\left(\frac fg\right)^{n-1}+\dots + a_n=0$ with coefficients $a_i\in A$. After clearing denominators and abusing notation we get $f^n+a_1f^{n-1}g+\cdots+a_ng^n=(x^2-yz)h$, where $f,g \in k[x,y]$ and $h,a_i\in k[x,y,z]$. Maybe I could take $\bmod z$ everything so that it simplifies as $f^n+a_1f^{n-1}g+\cdots+a_ng^n=hx^2$ but now with all polynomials are in 2 variables.
The next step naturally would be to say that $f$ and $g$ are not relatively prime. How? But I didn't assume any of that because the initial ring is not UFD.
How do I proceed?
 A: The solution described in the comments works out, but there is a more general and perhaps simpler argument for characteristic $\neq 2$. This can be found as Exercise II.6.4 in Hartshorne's algebraic geometry:
Let $k$ be a field of characteristic $\neq 2$ and let $f \in k[x_1, \dots, x_n]$
be a square free noncanstant polynomial. Let $A = k[x_1, \dots, x_n, z]/(z^2-f).$
As $f$ is square-free, the quotient field of $A$ is just $K = k(x_1, \dots,
x_n)[z]/(f-z^2)$ (the polynomial $f-z^2 \in k(x_1, \dots, x_n)[z]$ is
irreducible, thus $K$ is a field).
If now $\alpha = g + hz \in K$, then the minimal polynomial of $\alpha$ over $k(x_1, \dots, x_n)$ is
$$
    m_\alpha = (X-\alpha)(X+\alpha) = X^2 - 2gX + (g^2 - h^2f).
$$
This shows that if $\alpha \in K$
is integral over $A$, then (by Gauss's Lemma) $m_\alpha$ is the
minimal polynomial of $\alpha$ over $A$, and hence $2g, g^2-h^2f \in A$.
But now $g \in k[x_1, \dots, x_n]$, so $h^2f \in k[x_1, \dots, x_n]$.
Write $h = \frac pq$ (with $(p,q) = 1$). As polynomial rings are UFDs, $h^2f \in
k[x_1, \dots, x_n]$ now implies $q^2 \mid f$ which by square-freeness of $f$ gives
$q=1$. Now $\alpha \in A$, and we are done.
