Answer
No.
Work Step by Step
Solve for $y$:
$x^2-4y^2=1\\4y^2=x^2-1\\y^2=\frac{x^2-1}{4}\\y=\pm\frac{x^2-1}{4}.$
This relation is s not a function, because a distinct $x$-value corresponds to two different $y$-values.
For example, $(0, \frac{1}{4})$ and $(0, –\frac{1}{4}).$
