I'm having trouble finding the inverse of $y=\dfrac{\sin^2(x)}{x^2}$. 

So far what I have is 
$$\begin{align}
\frac{\sin^2(y)}{y^2} &= x \\[4pt]
\sin^2(y) &= xy^2 \\[4pt]
\sin(y) &= \sqrt{xy^2} = \sqrt{x}y \\[4pt]
y &= \sin^{-1}(\sqrt{x}y)
\end{align}$$

But I don't know how to take it any further than that. Does an inverse exist?