$$y=\frac{1}{\sqrt{x^{2}+2x}}$$
$x=\frac{\sqrt{y^{4}+y^{2}}}{y^{2}}-1$
$x=-\frac{\sqrt{y^{4}+y^{2}}}{y^{2}}-1\text{, }arg(\sqrt{\frac{1}{y^{2}}}y)<\pi \text{ and }y\neq 0$
$y=\left(x\left(x+2\right)\right)^{-\frac{1}{2}}$
$x\neq -2\text{ and }x\neq 0$
$x=\frac{\sqrt{y^{2}+1}}{y}-1$
$x=-\frac{\sqrt{y^{2}+1}}{y}-1\text{, }y>0$
$y=\frac{1}{\sqrt{x\left(x+2\right)}}$
$x>0\text{ or }x<-2$