Express $y$ from $\ln(x)+3\ln(y) = y$ i am finding a inverse function to $$y=\frac{e^\frac{x^3}{3}}{x^3}$$
First step:
$$x=\frac{e^\frac{y^3}{3}}{y^3}$$
Next: ... $$(xy^3)^3 = e^{{y}^3}$$
and now in this step i have no idea how can i express "y" 
$$\ln(x)+3\ln(y) = y$$
 A: Go back to the original:
$$y = x^{-3}e^{x^3/3}$$
Then: $$\frac{-1}{y}=-x^3e^{-x^3/3}$$
So: $$\frac{-1}{3y} = \frac{-x^3}{3}e^{-x^3/3}$$
This means that:
$$\frac{-x^3}{3} = W\left(\frac{-1}{3y}\right)$$
where $W$ is Lambert's $W$-function, with $W$ an inverse of $ze^z$.
So $$x=\left(-3W\left(\frac{-1}{3y}\right)\right)^{1/3}$$
You are not going to be able to do it without Lambert or some other non-elementary function.
A: Consider the function
$$
f(x)=\frac{e^{x^3/3}}{x^3}
$$
This function is defined for $x\ne0$. Also
$$
\lim_{x\to-\infty}f(x)=0,\quad
\lim_{x\to0^-}f(x)=-\infty,\quad
\lim_{x\to0^+}f(x)=\infty,\quad
\lim_{x\to\infty}f(x)=\infty
$$
so the function has no “global” inverse.
Its derivative is
$$
f'(x)=\frac{x^5e^{x^3/3}-3x^2e^{x^3/3}}{x^6}=\frac{x^3-3}{x^4}e^{x^3/3}
$$
which vanishes for $x=\sqrt[3]{3}$ and is negative for $x<\sqrt[3]{3}$ (but $x\ne0$, of course) and positive for $x>\sqrt[3]{3}$.
So the function is decreasing in the intervals $(-\infty,0)$ and $(0,\sqrt[3]{3}]$; it is increasing in the interval $[\sqrt[3]{3},\infty)$.
How can we find the range? We know that the interval $(-\infty,0)$ is contained in the range (consider the $(-\infty,0)$ part of the domain). Since
$$
f(\sqrt[3]{3})=\frac{e}{3}
$$
we know that the range corresponding to the part $(0,\infty)$ of the domain is $[e/3,\infty)$.
Putting things together, the range is
$$
(-\infty,0)\cup[e/3,\infty)
$$
The restriction of $f$ to each of those intervals has an inverse function. Which is, well, the inverse function: there's no “explicit” form for it. But it's not necessary to find an inverse function, in order to determine the range.
