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Here is a proof using a bit of commutative algebra and algebraic geometry, elaborating on comments by @abx and @ChrisWuthrich. The ring $$R=\mathbb{C}[x,y]/(x^3+y^3-1)$$ is a Dedekind domain, because it is the coordinate ring of the affine elliptic curve $$E=\{(x,y)\in\mathbb{C}^2:x^3+y^3=1\}.$$ As a result, if $R$ is a UFD, then it is also a PID, hence the Picard group $\mathrm{Pic}(E)$ is trivial. We shall show that this is not the case.

Consider the projective closure of $E$: $$\tilde E=\{(x:y:z)\in\mathbf{P}(\mathbb{C}^3):x^3+y^3=z^3\}.$$ We have an embedding $f:E\to\tilde E$ given by $(x,y)\mapsto(x,y,1)$, and in fact $\tilde E$ is the disjoint union of $f(E)$ and the set of points at infinity, $$S=\{(-1:1:0), (-1:e^{2\pi i/3}:0), (-1:e^{4\pi i/3}:0)\}.$$ Restricting divisors of $\tilde E$ to $f(E)=\tilde E\setminus S$ gives rise to an exact sequence $$0\to\mathrm{Pic}(\tilde E,S)\to\mathrm{Pic}(\tilde E)\to\mathrm{Pic}(E)\to 0,$$ where $\mathrm{Pic}(\tilde E,S)$ is the group of divisors of $\tilde E$ supported on $S$ modulo principal divisors of $\tilde E$ supported on $S$. Note that $\mathrm{Pic}(\tilde E,S)$ is a finitely generated abelian group, while $\mathrm{Pic}(\tilde E)$ is isomorphic to $\mathbb{Z}\times\tilde E$ if we regard $\tilde E$ as an abelian group in the usual way. Hence $\mathrm{Pic}(E)$ has cardinality continuum, and $R$ is not a UFD.

Remark 1. The field $L=\mathbb{C}(x,y)/(x^3+y^3-1)$ is a cubic Galois extension of $K=\mathbb{C}(x)$. The ideal $(x)$ of $R$ factors into prime ideals as $$(x)=(x,y-1)(x,y-e^{2\pi i/3})(x,y-e^{4\pi i/3}),$$ and the three factors are cyclically permuted by $\mathrm{Gal}(L/K)$. In fact these factors are not principal as the answer of user516477 shows below. Equivalently, $x$ is irreducible (but not prime) in $R$.

Remark 2. Note that $x+y$, $x+e^{2\pi/3}y$, $x+e^{4\pi/3}y$ are units in $R$ as their product is $x^3+y^3=1$. It can be derived from the calculations of user516477 that these three units and $\mathbb{C}^\times$ generate $R^\times$. It follows that $\mathrm{Pic}(\tilde E,S)$ is isomorphic to $\mathbb{Z}\times(\mathbb{Z}/3)$, and $\mathrm{Pic}(E)$ is isomorphic to $\mathbb{Z}\times(\tilde E/G)$ , where $G$ is a certain 3-element subgroup of $\tilde E$.

Here is a proof using a bit of commutative algebra and algebraic geometry, elaborating on comments by @abx and @ChrisWuthrich. The ring $$R=\mathbb{C}[x,y]/(x^3+y^3-1)$$ is a Dedekind domain, because it is the coordinate ring of the affine elliptic curve $$E=\{(x,y)\in\mathbb{C}^2:x^3+y^3=1\}.$$ As a result, if $R$ is a UFD, then it is also a PID, hence the Picard group $\mathrm{Pic}(E)$ is trivial. We shall show that this is not the case.

Consider the projective closure of $E$: $$\tilde E=\{(x:y:z)\in\mathbf{P}(\mathbb{C}^3):x^3+y^3=z^3\}.$$ We have an embedding $f:E\to\tilde E$ given by $(x,y)\mapsto(x,y,1)$, and in fact $\tilde E$ is the disjoint union of $f(E)$ and the set of points at infinity, $$S=\{(-1:1:0), (-1:e^{2\pi i/3}:0), (-1:e^{4\pi i/3}:0)\}.$$ Restricting divisors of $\tilde E$ to $f(E)=\tilde E\setminus S$ gives rise to an exact sequence $$0\to\mathrm{Pic}(\tilde E,S)\to\mathrm{Pic}(\tilde E)\to\mathrm{Pic}(E)\to 0,$$ where $\mathrm{Pic}(\tilde E,S)$ is the group of divisors of $\tilde E$ supported on $S$ modulo principal divisors of $\tilde E$ supported on $S$. Note that $\mathrm{Pic}(\tilde E,S)$ is a finitely generated abelian group, while $\mathrm{Pic}(\tilde E)$ is isomorphic to $\mathbb{Z}\times\tilde E$ if we regard $\tilde E$ as an abelian group in the usual way. Hence $\mathrm{Pic}(E)$ has cardinality continuum, and $R$ is not a UFD.

Remark 1. The field $L=\mathbb{C}(x,y)/(x^3+y^3-1)$ is a cubic Galois extension of $K=\mathbb{C}(x)$. The ideal $(x)$ of $R$ factors into prime ideals as $$(x)=(x,y-1)(x,y-e^{2\pi i/3})(x,y-e^{4\pi i/3}),$$ and the three factors are cyclically permuted by $\mathrm{Gal}(L/K)$. In fact these factors are not principal as the answer of user516477 shows below. Equivalently, $x$ is irreducible (but not prime) in $R$.

Remark 2. Note that $x+y$, $x+e^{2\pi/3}y$, $x+e^{4\pi/3}y$ are units in $R$ as their product is $x^3+y^3=1$. It can be derived from the calculations of user516477 that these three units and $\mathbb{C}^\times$ generate $R^\times$. It follows that $\mathrm{Pic}(\tilde E,S)$ is isomorphic to $\mathbb{Z}\times(\mathbb{Z}/3)$, and $\mathrm{Pic}(E)$ is isomorphic to $\tilde E$ divided by a certain 3-element subgroup.

Here is a proof using a bit of commutative algebra and algebraic geometry, elaborating on comments by @abx and @ChrisWuthrich. The ring $$R=\mathbb{C}[x,y]/(x^3+y^3-1)$$ is a Dedekind domain, because it is the coordinate ring of the affine elliptic curve $$E=\{(x,y)\in\mathbb{C}^2:x^3+y^3=1\}.$$ As a result, if $R$ is a UFD, then it is also a PID, hence the Picard group $\mathrm{Pic}(E)$ is trivial. We shall show that this is not the case.

Consider the projective closure of $E$: $$\tilde E=\{(x:y:z)\in\mathbf{P}(\mathbb{C}^3):x^3+y^3=z^3\}.$$ We have an embedding $f:E\to\tilde E$ given by $(x,y)\mapsto(x,y,1)$, and in fact $\tilde E$ is the disjoint union of $f(E)$ and the set of points at infinity, $$S=\{(-1:1:0), (-1:e^{2\pi i/3}:0), (-1:e^{4\pi i/3}:0)\}.$$ Restricting divisors of $\tilde E$ to $f(E)=\tilde E\setminus S$ gives rise to an exact sequence $$0\to\mathrm{Pic}(\tilde E,S)\to\mathrm{Pic}(\tilde E)\to\mathrm{Pic}(E)\to 0,$$ where $\mathrm{Pic}(\tilde E,S)$ is the group of divisors of $\tilde E$ supported on $S$ modulo principal divisors of $\tilde E$ supported on $S$. Note that $\mathrm{Pic}(\tilde E,S)$ embeds into $\mathbb{Z}^3$, while $\mathrm{Pic}(\tilde E)$ is isomorphic to $\mathbb{Z}\times\tilde E$ if we regard $\tilde E$ as an abelian group in the usual way. Hence $\mathrm{Pic}(E)$ has cardinality continuum, and $R$ is not a UFD.

Remark. The field $L=\mathbb{C}(x,y)/(x^3+y^3-1)$ is a cubic Galois extension of $K=\mathbb{C}(x)$. The ideal $(x)$ of $R$ factors into prime ideals as $$(x)=(x,y-1)(x,y-e^{2\pi i/3})(x,y-e^{4\pi i/3}),$$ and the three factors are cyclically permuted by $\mathrm{Gal}(L/K)$. In fact these factors are not principal as the answer of user516477 shows below. Equivalently, $x$ is irreducible (but not prime) in $R$.

Here is a proof using a bit of commutative algebra and algebraic geometry, elaborating on comments by @abx and @ChrisWuthrich. The ring $$R=\mathbb{C}[x,y]/(x^3+y^3-1)$$ is a Dedekind domain, because it is the coordinate ring of the affine elliptic curve $$E=\{(x,y)\in\mathbb{C}^2:x^3+y^3=1\}.$$ As a result, if $R$ is a UFD, then it is also a PID, hence the Picard group $\mathrm{Pic}(E)$ is trivial. We shall show that this is not the case.

Consider the projective closure of $E$: $$\tilde E=\{(x:y:z)\in\mathbf{P}(\mathbb{C}^3):x^3+y^3=z^3\}.$$ We have an embedding $f:E\to\tilde E$ given by $(x,y)\mapsto(x,y,1)$, and in fact $\tilde E$ is the disjoint union of $f(E)$ and the set of points at infinity, $$S=\{(-1:1:0), (-1:e^{2\pi i/3}:0), (-1:e^{4\pi i/3}:0)\}.$$ Restricting divisors of $\tilde E$ to $f(E)=\tilde E\setminus S$ gives rise to an exact sequence $$0\to\mathrm{Pic}(\tilde E,S)\to\mathrm{Pic}(\tilde E)\to\mathrm{Pic}(E)\to 0,$$ where $\mathrm{Pic}(\tilde E,S)$ is the group of divisors of $\tilde E$ supported on $S$ modulo principal divisors of $\tilde E$ supported on $S$. Note that $\mathrm{Pic}(\tilde E,S)$ is a finitely generated abelian group, while $\mathrm{Pic}(\tilde E)$ is isomorphic to $\mathbb{Z}\times\tilde E$ if we regard $\tilde E$ as an abelian group in the usual way. Hence $\mathrm{Pic}(E)$ has cardinality continuum, and $R$ is not a UFD.

Remark 1. The field $L=\mathbb{C}(x,y)/(x^3+y^3-1)$ is a cubic Galois extension of $K=\mathbb{C}(x)$. The ideal $(x)$ of $R$ factors into prime ideals as $$(x)=(x,y-1)(x,y-e^{2\pi i/3})(x,y-e^{4\pi i/3}),$$ and the three factors are cyclically permuted by $\mathrm{Gal}(L/K)$. In fact these factors are not principal as the answer of user516477 shows below. Equivalently, $x$ is irreducible (but not prime) in $R$.

Remark 2. Note that $x+y$, $x+e^{2\pi/3}y$, $x+e^{4\pi/3}y$ are units in $R$ as their product is $x^3+y^3=1$. It can be derived from the calculations of user516477 that these three units and $\mathbb{C}^\times$ generate $R^\times$. It follows that $\mathrm{Pic}(\tilde E,S)$ is isomorphic to $\mathbb{Z}\times(\mathbb{Z}/3)$, and $\mathrm{Pic}(E)$ is isomorphic to $\mathbb{Z}\times(\tilde E/G)$ , where $G$ is a certain 3-element subgroup of $\tilde E$.

For a short and rather elementary proof, scroll down to the Added section of this post.

Here is a proof using a bit of commutative algebra and algebraic geometry, elaborating on comments by @abx and @ChrisWuthrich. The ring $$R=\mathbb{C}[x,y]/(x^3+y^3-1)$$ is a Dedekind domain, because it is the coordinate ring of the affine elliptic curve $$E=\{(x,y)\in\mathbb{C}^2:x^3+y^3=1\}.$$ As a result, if $R$ is a UFD, then it is also a PID, hence the Picard group $\mathrm{Pic}(E)$ is trivial. We shall show that this is not the case.

Consider the projective closure of $E$: $$\tilde E=\{(x:y:z)\in\mathbf{P}(\mathbb{C}^3):x^3+y^3=z^3\}.$$ We have an embedding $f:E\to\tilde E$ given by $(x,y)\mapsto(x,y,1)$, and in fact $\tilde E$ is the disjoint union of $f(E)$ and the set of points at infinity, $$S=\{(-1:1:0), (-1:e^{2\pi i/3}:0), (-1:e^{4\pi i/3}:0)\}.$$ Restricting divisors of $\tilde E$ to $f(E)=\tilde E\setminus S$ gives rise to an exact sequence $$0\to\mathrm{Pic}(\tilde E,S)\to\mathrm{Pic}(\tilde E)\to\mathrm{Pic}(E)\to 0,$$ where $\mathrm{Pic}(\tilde E,S)$ is the group of divisors of $\tilde E$ supported on $S$ modulo principal divisors of $\tilde E$ supported on $S$. Note that $\mathrm{Pic}(\tilde E,S)$ embeds into $\mathbb{Z}^3$, while $\mathrm{Pic}(\tilde E)$ is isomorphic to $\mathbb{Z}\times\tilde E$ if we regard $\tilde E$ as an abelian group in the usual way. Hence $\mathrm{Pic}(E)$ has cardinality continuum, and $R$ is not a UFD.

Remark. The field $L=\mathbb{C}(x,y)/(x^3+y^3-1)$ is a cubic Galois extension of $K=\mathbb{C}(x)$. The ideal $(x)$ of $R$ factors into prime ideals as $$(x)=(x,y-1)(x,y-e^{2\pi i/3})(x,y-e^{4\pi i/3}),$$ and the three factors are cyclically permuted by $\mathrm{Gal}(L/K)$. In fact these factors are not principal as the answer of user516477 shows below. Equivalently, $x$ is irreducible (but not prime) in $R$.

Here is a proof using a bit of commutative algebra and algebraic geometry, elaborating on comments by @abx and @ChrisWuthrich. The ring $$R=\mathbb{C}[x,y]/(x^3+y^3-1)$$ is a Dedekind domain, because it is the coordinate ring of the affine elliptic curve $$E=\{(x,y)\in\mathbb{C}^2:x^3+y^3=1\}.$$ As a result, if $R$ is a UFD, then it is also a PID, hence the Picard group $\mathrm{Pic}(E)$ is trivial. We shall show that this is not the case.

Consider the projective closure of $E$: $$\tilde E=\{(x:y:z)\in\mathbf{P}(\mathbb{C}^3):x^3+y^3=z^3\}.$$ We have an embedding $f:E\to\tilde E$ given by $(x,y)\mapsto(x,y,1)$, and in fact $\tilde E$ is the disjoint union of $f(E)$ and the set of points at infinity, $$S=\{(-1:1:0), (-1:e^{2\pi i/3}:0), (-1:e^{4\pi i/3}:0)\}.$$ Restricting divisors of $\tilde E$ to $f(E)=\tilde E\setminus S$ gives rise to an exact sequence $$0\to\mathrm{Pic}(\tilde E,S)\to\mathrm{Pic}(\tilde E)\to\mathrm{Pic}(E)\to 0,$$ where $\mathrm{Pic}(\tilde E,S)$ is the group of divisors of $\tilde E$ supported on $S$ modulo principal divisors of $\tilde E$ supported on $S$. Note that $\mathrm{Pic}(\tilde E,S)$ embeds into $\mathbb{Z}^3$, while $\mathrm{Pic}(\tilde E)$ is isomorphic to $\mathbb{Z}\times\tilde E$ if we regard $\tilde E$ as an abelian group in the usual way. Hence $\mathrm{Pic}(E)$ has cardinality continuum, and $R$ is not a UFD.

Remark. The field $L=\mathbb{C}(x,y)/(x^3+y^3-1)$ is a cubic Galois extension of $K=\mathbb{C}(x)$. The ideal $(x)$ of $R$ factors into prime ideals as $$(x)=(x,y-1)(x,y-e^{2\pi i/3})(x,y-e^{4\pi i/3}),$$ and the three factors are cyclically permuted by $\mathrm{Gal}(L/K)$. In fact these factors are not principal as the answer of user516477 shows below. Equivalently, $x$ is irreducible (but not prime) in $R$.