I'm trying to invert the following function: $f(x) = \frac{1}{2}x^2\ln{x}-\frac{1}{4}x^2$ for all $x>1$.
I compared the expression to $y$ in order to get $x$, but I don't know how to proceed.
It will be great to understand what should be the process.
If you are allowed to use the Lambert $W$ function, then the procedure is roughly (there are a couple of domain considerations that must be taken into account): \begin{align} 4y&=(-1+2\ln x)x^2=(-1+\ln x^2)x^2\\ \frac{4y}{e}&=\left(\frac xe\right)^2\ln\left(\frac xe\right)^2&\text{call }s:=\frac{x^2}{e^2},\ t:=\frac{4y}{e}\\\frac ts&=\ln s\\ e^{t/s}&=s\\\frac1se^{t/s}&=1\\\frac ts&=W(t)\\\frac{x^2}{e^2}=s&=\frac{t}{W(t)}=\frac{4y}{e\cdot W(4y/e)}\\x&=2\sqrt{\frac{ey}{W(4y/e)}}\end{align}