Note that $\frac{x^2}{9}=\left(\frac{x}{3}\right)^2$. Also note that you can multiply both sides of your equation by $y^2$ to get just one $y$ term, and that the cube root is the same as the $\frac{1}{3}$ power.
You could, as lhf indicates, avoid ever breaking up the $y$ terms in the first place. Technically this would be preferred, because when you divided by $y$ you made the assumption that $y\neq 0$, which is not always true.
Beware of squaring when solving equations in general, because it can lead to extraneous solutions. In this case it is a valid step, but for a silly example consider trying to solve the equation $\sqrt{y}=-1$ by squaring both sides.
A final remark. The original function has an implicit domain, $x\geq 0$. Your inverse function has a formula definition that does not appear to require any domain restriction, but it is misleading to say that $f^{-1}(x)=\left(\frac{x}{3}\right)^{2/3}$ without mentioning that this is only valid when $x\geq 0$. This domain restriction arises from the fact that the range of $f$ is $[0,\infty)$. The function $h(x)=\left(\frac{x}{3}\right)^{2/3}$ on the whole real line is not invertible (it fails the horizontal line test), whereas if $g(x)=-f(x)=-3x\sqrt{x}$, then the inverse function would be $g^{-1}(x)=\left(\frac{x}{3}\right)^{2/3}$ on $(-\infty,0]$. Looking at the graphs of these functions and keeping in mind how inverses are reflected across the line $y=x$ (resulting from switching $x$ and $y$) should make it clearer what is going on.
