$\left\{\begin{matrix}x=\frac{\left(\frac{\sqrt{1-8y^{2}}+1}{y}\right)^{2}}{4}\text{; }x=\frac{\left(\frac{\sqrt{1-8y^{2}}-1}{y}\right)^{2}}{4}\text{, }&y>0\text{ and }y\leq \frac{\sqrt{2}}{4}\\x=0\text{, }&y=0\end{matrix}\right.$

$y=\frac{\sqrt{x}}{x+2}$
$x\geq 0$