Assume $k$ is a alg closed field. Then it is easy to check that the morphism $\phi: \mathbb{P}^1 \to \mathbb{P}^2, (x_0:x_1) \mapsto (x_0^2: x_0x_1:x_1^2)$ induces an isomorphism between $\mathbb{P}^1$ and homogeneous curve $C=V_+(y_0^2-y_1^2 +_2^2)$ as follows:
The image of $ \mathbb{P}^1 $ by $\phi$ is $V_+(XZ-Y^2)$ and if we make substitution $X:= y_0+iy_2, Z:= y_0-iy_2$ and $Y:=y_1$ we obtain the equation for $C$.
This curve has homogeneous coordinate ring $k[y_0,y_1,y_2]/(y_0^2-y_1^2 +y_2^2)$. We know that since $\mathbb{P}^1$ is regular every stalk $\mathcal{\mathbb{P}^1,p}$ is a regular local ring = a UFD since $\mathbb{P}^1$ integral of dimension $1$.
Since $\mathbb{P}^1 \cong C$ every stalk of $C$ at arbitrary closed point has to be also a UFD. Unfortunately, I not see why this is true if we forget $\mathbb{P}^1 \cong C$ and try to do the calculation of $\mathcal{O}_{C,c}$ by hand. Consider the affine $D_+(y_0)$-chart, then we have to prove that every localization of the affine ring $k[y,z]/(1-y^2 +z^2)$ at any maximal prime of $k[y,z]/(1-y^2 +z^2)$ (ie every prime exept $(1-y^2 +z^2)$ is a UFD. How can this claim be checked?