Let $A$ be $ℂ[x, y]$ and let $I$ be the given ideal in $A$.
Let $J ≔ ⟨x + 3y - 1⟩$, which is an ideal in $A$ with $A ⊇ I ⊇ J$.
By the third isomorphism theorem we have
$$
A / I ≅ (A / J) / (I / J) \,.
$$
The quotient $A / J$ is easy to determine:
modding out the polynomial $x + 3y - 1$ identifies $x$ with $1 - 3y$.
More explicitly, we have the isomorphism
$$
φ \colon A / J ≅ ℂ[y] \,, \quad
[x] \mapsto 1 - 3y \, \quad
[y] \mapsto y \,.
\tag{1}
$$
Under the isomorphism $φ$ the ideal $I / J$ of $A / J$ corresponds to an ideal $K$ in $ℂ[y]$, and the isomorphism $φ$ descends to an isomorphism
$$
(A / J) / (I / J) ≅ ℂ[y] / K \,.
$$
The ideal $I / J$ is generated by the two residue classes $[x^2 + 2xy + y^2 - 529]$ and $[x + 3y - 1]$, whence the ideal $K$ is generated by $φ([x^2 + 2xy + y^2 - 529])$ and $φ([x + 3y - 1])$.
The second generator of $K$ is $0$ because already $[x + 3y - 1] = 0$ in $A / J$ (after all, we modded out this generator when we formed $A / J$).
The first generator of $K$ is
$$
(1 - 3y)^2 + 2(1 - 3y)y + y^2 - 529
=
4 y^2 - 4 y - 528
=
4 ⋅ (y^2 - y - 132) \,.
$$
The factor $4$ is invertible in $ℂ[y]$, whence $K$ is generated by the single polynomial $y^2 - y - 132$.
We have the decomposition
$$
y^2 - y - 132
=
(y - 12) (y + 11) \,.
$$
It now follows from the Chinese Remainder Theorem (for principal ideal domains) that
\begin{align*}
ℂ[y] / K
&=
ℂ[y] / ⟨y^2 - y - 132⟩ \\
&=
ℂ[y] / ⟨(y - 12)(y + 11)⟩ \\
&≅
ℂ[y] / ⟨y - 12⟩ × ℂ[y] / ⟨y + 11⟩ \tag{2} \\
&≅
ℂ × ℂ \,.
\end{align*}
We use here that
$$
ℂ[x] / ⟨x - a⟩ ≅ ℂ \,, \quad
[p] \mapsto p(a) \tag{3}
$$
for every $a ∈ ℂ$.
If you wish to write down an explicit isomorphism between $A/I$ and $ℂ × ℂ$ then you’ll need to trace the images of $[x]$ and $[y]$ under the above sequence of isomorphisms
\begin{align*}
A / I
&≅ (A / J) / (I / J) \\
&≅ ℂ[y] / K \\
&≅ ℂ[y] / ⟨y - 12⟩ × ℂ[y] / ⟨y + 11⟩ \\
&≅ ℂ × ℂ \,.
\end{align*}
We already have explicit formulas for the steps $(1)$ and $(3)$, and the isomorphism $(2)$ is given by $[p] \mapsto ([p], [p])$.
We thus find that
$$
[x]
\mapsto [[x]]
\mapsto [1 - 3y]
\mapsto ([1 - 3y], [1 - 3y])
\mapsto (1 - 3 ⋅ 12, 1 - 3 ⋅ (-11))
= (-35, 34)
$$
and
$$
[y]
\mapsto [[y]]
\mapsto [y]
\mapsto ([y], [y])
\mapsto (12, -11) \,.
$$
The overall isomorphism between $A / I$ and $ℂ × ℂ$ is thus given by sending $[x]$ to $(-35, 34)$ and $[y]$ to $(12, -11)$.
