Answer
$y$ is NOT a function of $x$.
Work Step by Step
Solve for $y$ in terms of $x$ to obtain:
$\begin{array}{ccc}
&x &= &y^2
\\&\pm\sqrt{x} &= &y
\end{array}$
The last equation above will give two different values of $y$ for each value of $x$.
Thus, the equation does not represent $y$ as a function of $x$.
