Answer
The equation DOES NOT represent $y$ as a function of $x$.
Work Step by Step
Solve for $y$ in terms of $x$ to have:
$2x^2-3=4y^2
\\\frac{2x^2-3}{4}=y^2
\\\pm \sqrt{\dfrac{2x^2-3}{4}}=y
\\y=\pm \sqrt{\dfrac{2x^2-3}{4}}$
The last equation is a rule that gives two values of $y$ for some values of $x$ (e.g., x=4, x=5).
Thus, the equation does not define $y$ as a function of $x$.
