If $y\left(x\right)=x^x,x>0$, then $y^{"}\left(2\right)-2y^{\text{'}}\left(2\right)$ is equal to :

Given, $y=x^x$

$\Rightarrow \ln \left(y\right)=x\ln \left(x\right)$

Differentiate both sides w.r.t. $x$

$\frac{1}{y}.y\text{'}=1+\ln \left(x\right)$

$\Rightarrow y\text{'}=y\left(1+lnx\right) .......\left(1\right)$

$\Rightarrow y\text{'}=x^x\left(1+lnx\right)$

at $x=2\Rightarrow y=4$

So, $y\text{'}\left(2\right)=4\left(1+ln2\right) .......\left(2\right)$

Now differentiate the equation $\left(1\right)$,

$y"=y\text{'}\left(1+\ln x\right)+\frac{y}{x}$

$\Rightarrow y"\left(2\right)=y\text{'}\left(1+\ln 2\right)+2$

$\Rightarrow y"\left(2\right)-y\text{'}\left(2\right)=y\text{'}\left(\ln 2\right)+2$

$\Rightarrow y"\left(2\right)-2y\text{'}\left(2\right)=\left(\ln 2-1\right)y\text{'}\left(2\right)+2$

$=4(\ln 2-1)(\ln 2+1)+2$

$=4{\left(\ln 2\right)}^2-2$