Answer
$y$ is NOT a function of $x$.
Work Step by Step
Solve for $y$ in terms of $x$ to obtain:
$\begin{array}{ccc}
&2x-4y^2 &= &3
\\&-4y^2 &= &3-2x
\\&\dfrac{-4y^2}{-4} &= &\dfrac{3-2x}{-4}
\\&y^2 &= &-\dfrac{3}{4}+\dfrac{1}{2}x
\\&y &= &\pm \sqrt{-\dfrac{3}{4}+\dfrac{1}{2}x}
\end{array}$
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 $x$.
