Answer
No.
Work Step by Step
Solve for $y$:
$2x^2+3y^2=1\\3y^2=1-2x^2\\y^2=\frac{1-2x^2}{3}\\y=\pm\frac{1-2x^2}{3}.$
This relation is s not a function, because a distinct $x$-value corresponds to two different $y$-values.
For example, $(0, \frac{1}{3})$ and $(0, –\frac{1}{3}).$
