Solution: Define a map
\[
f: \mathbb{R} ^2 \to \mathbb{R} , ~ (x,y) \mapsto x + y^2.
\]
Note that the map $f$ is continuous. Also, if $\left( x_1, y_1 \right) \sim \left( x_2, y_2 \right) $, then
\[
x_1 + y_1^2 = x_2 + y_2^2 \implies f\left( x_1,y_1 \right) = f\left( x_2, y_2 \right)
\]
and hence $f$ preserves the equivalence relation. Therefore, the following theorem implies that there is a continuous map from the quotient space to $\mathbb{R} $.
Let $X$ and $Y$ be two topological space and let $f : X \to Y$ be a continuous map. Let $\sim $ be the equivalence relation on $X$ defined by $x_1 \sim x_2$ if and only if $f\left( x_1 \right) = f\left( x_2 \right) $. Then there exists a continuous function $g : X / \mathord{\sim } \to Y$ such that $f = g\circ \pi$.
So we got a continuos map $g : \mathbb{R} ^2 / \mathord{\sim } \to \mathbb{R} $ defined by $g[(x,y)] \mapsto x + y^2 $. Define a map
\[
h : \mathbb{R} \to \mathbb{R} ^2 / \mathord{\sim },~ x \mapsto [(x,0)].
\]
Note that $h$ is defined as the composition of maps $h_1 : \mathbb{R} \to \mathbb{R} ^2, ~ x \mapsto (x,0)$ and the quotient map. Therefore, it is continuous. Now we only need to show that $g \circ h$ and $h\circ g$ are identity.
\begin{align*}
(g \circ h)(x) & = g([x,0]) = x \\
(h \circ g)([x,y]) & = h\left( x + y^2 \right) = \left[ (x + y^2, 0) \right] = [(x,y)].
\end{align*}
Therefore, the quotient space is homeomorphic to $\mathbb{R} $.