Clearly, the domain of $y(x)$ is all real $x$. If one applies derivatives to $y(x)$, the critical points at $y'(x)= 0$ occur at $x = 0, \pm \sqrt{7/2}$, which in turn yield the extreme values of $\frac{1}{2}$ and $\frac{\sqrt{15}}{8}$ respectively. Hence, the range of $y(x)$ is $[\frac{\sqrt{15}}{8}, \frac{1}{2}]$.