Let $g(x)=(x-a)(x-b)(x-c)$ with $a\neq b\neq c$. Then $\mathbb{C}[x,y]/\langle y^2-g(x) \rangle$ is integrally closed in its fraction field.
I manage to do with the following theorem:
A smooth function $f$ is non-singular if and only if $\mathbb{C}[x,y]/\langle f(x,y) \rangle$ is integrally closed in its fraction field.
But how will I do without using this? Like an example has done here.