Solving an Equation for y in terms of x How is it possible to solve the equation:
$$ x = \frac{2-y}{y} \sqrt{1-y^2} $$
for $ y $ in terms of $ x $?
I know that it is possible to do it, since WolframAlpha gives a complicated answer but entirely composed of elementary functions. However, I am stumped on how to even start.
Thanks in advance!

This is not an answer but it is too long for a comment.
As shown in previous comments and answers, $$f(y) = \frac{2-y}{y} \sqrt{1-y^2}$$ is only defined for $-1\leq y\lt 0$ and $0\lt y\leq 1$. If $x>0$, for $f(y)=x$ ,there is only one solution corresponding to $y>0$ and if $x<0$ there is only one solution corresponding to $y<0$. All possible remaining roots   were introduced by squaring.
For a given large value of $x$, using Taylor series, $f(y)$ can be approximated by $$f(y)=\frac{2}{y}-1-y+O\left(y^2\right)$$ which give solutions $$y_{\pm}=\frac{1}{2} \left(-x-1\pm\sqrt{x^2+2 x+9}\right)$$ ($y_+$ corresponding to $x>0$ and $y_-$ corresponding to $x<0$).
For example, for $x=10$, the approximation gives $y\approx 0.178908$ while the exact solution should be $y\approx 0.179140$; for $x=-10$, the approximation gives $y\approx -0.216991$ while the exact solution should be $y\approx -0.216388$.
For a more rigorous numerical solution, Newton method could be used since we already have good initial values.
A better approximation of $f(y)$ is given by a Pade approximant $$f(y)\approx\frac{2-\frac{3 y^2}{2}}{y+\frac{y^2}{2}}$$ which would approximate the solution as $$y_{\pm}=\frac{-x\pm\sqrt{x^2+4 x+12}}{x+3}$$ For the above example, the aproximations are almost identical to the solutions.
