$$y=\frac{1}{x^{2}}+2$$
$x=-i\left(2-y\right)^{-\frac{1}{2}}$
$x=i\left(2-y\right)^{-\frac{1}{2}}\text{, }y\neq 2$
$x=\sqrt{-\frac{1}{2-y}}$
$x=-\sqrt{-\frac{1}{2-y}}\text{, }y>2$
$y=2+\frac{1}{x^{2}}$
$x\neq 0$