Normal if and only if is UFD 
If we consider $f \in \mathbb{C}[x,y]$ an irreducible polynomial, then it is true that the domain $ \mathbb{C}[x,y]/(f)$ is normal iff it is UFD? 

I think this is false. I was trying to prove that $\mathbb{C}[x,y]/(x^2-y)$ isn't a UFD using the equality $x^2 = y$, although it is normal, because $x^2 - y$ is regular curve.
 A: This is untrue.  Being a UFD implies normal but not conversely.  It is true that normal and Cl(R) = 0 is equivalent to UFD.
For a counterexample, let's look for a normal variety whose class ring of Weil divisors is nonzero.  I know one:
$$k[x,y,z]/(xy - z^2)$$
I know it is normal because it is Cohen Macaulay (complete intersection) and it is regular in codimension one (i.e. its singularities are in codimension $\geq$ 2).  It's class group of Weil divisors is $\mathbb{Z}/2$ but it's easier to just see directly that it's not a UFD: $z^2 = xy$.
Some of the above is probably too much theory for this question, but I find it easier to think about these things "geometrically."
For a reference, see Hartshorne's Algebraic Geometry, the section on divisors in chapter 2.  This will do a better job than me on the finer details.
Edit: Apologies: in your situation normal is equivalent to UFD.
A normal dimension one variety is regular.  You can probably find this result in Hartshorne, but it's also on Wikipedia:
http://en.wikipedia.org/wiki/Normal_scheme#The_normalization
Regular local rings are UFDs by Auslander-Buchsbaum, and I think a ring that is locally a UFD is a UFD, which you can check by using one of these equivalent definitions (http://en.wikipedia.org/wiki/Unique_factorization_domain#Equivalent_conditions_for_a_ring_to_be_a_UFD).
Edit 2: Normal dimension one varieties are regular using Serre's characterization of normal varieties as (i) regular in codimension one and (ii) some condition which is satisfied by Cohen-Macaulay rings, in particular for complete intersections.  A reference is http://stacks.math.columbia.edu/tag/033P and the conditions are R1 and S2 in their notation.  An integral (reduced) dimension one variety regular in codimension one is regular.
A: This is indeed not true: an explicit counterexample is $R = \mathbb{C}[x,y]/(y^2 - x^3 + x)$. This is a Dedekind domain that is not a UFD, as $(x,y)$ is a height $1$ prime that is not principal (although it is so locally). To see that $y^2 - x^3 + x$ is irreducible, one can use Eisenstein's criterion at the prime element $x$. 
Geometrically, rings of the form $\mathbb{C}[x,y]/(f)$, $f$ irreducible, correspond to affine plane (integral) curves in $\mathbb{A}^2_{\mathbb{C}}$. Requiring $\mathbb{C}[x,y]/(f)$ to be normal is equivalent to requiring the curve to be nonsingular, since as pointed out in the other answer, $\mathbb{C}[x,y]/(f)$ is always Cohen-Macaulay. However, it turns out that the coordinate ring of a nonsingular affine plane curve is a UFD iff the curve is rational, i.e. has genus $0$ (see e.g. Corollary 3.23 here). The example above is in some sense the simplest counterexample, being the standard example of an affine elliptic curve, which has genus $1$.
