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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.

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    have you had a look at the graph of $f(x)$? this function hasn't an inverse – 2017-01-22
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    It is not 1-1 function in $R$ . $f(0) \to 0 and f(1.649)=0 $ because $$\frac14x^2(2lnx-1)=0 \to x=0 ,lnx=\frac12 \to x=1.649$$ – 2017-01-22
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    You're right, I meant for all numbers bigger then 1. – 2017-01-22
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    It's not a function if you don't specify a set for $x$. Oh wait you had specified $x>1$ – 2017-01-22

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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}

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    Didn't learned it... How can I do it by the regular method? – 2017-01-22
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    @kurt If by "regular method" you mean "with elementary functions like $\sin$, $\ln$, $\exp$, $\arccos$ and polynomials", then you cannot. – 2017-01-22