If $y=x^{\left(x^x\right)},$  then $\frac{dy}{dx}$ is

$y_1=x^x$ $\Rightarrow$ $\log y_1=x\log x$

$\frac{1}{y_1}\frac{dy}{dx}=x.\frac{1}{x}+\log x\Rightarrow \frac{dy}{dx}=y_1\left[1+\log x\right]\Rightarrow x^x\left(1+\log x\right)$

$y=x^{\left(x^x\right)}$

$\log y={\log }_e^{x^{{\left(x\right)}^x}}$

$\log y=x^x{\log }_e^x$

$\frac{1}{y}\frac{dy}{dx}=x^x\frac{d}{dx}\left({\log }_e^x\right)+\left({\log }_e^x\right)\frac{d}{dx}\left(x^x\right)$

$\frac{dy}{dx}=y\left[x^x.\frac{d}{dx}\left(\frac{\log x}{\log e}\right)+{\log }_e^x{x}^x\left(1+{\log }_e^x\right)\right]$

$=y\left[x^x.\frac{1}{x\log e}+{\log }_e^x\left[{\log }_e^e+{\log }_e^x\right]x^x\right]$

$=y\left[x^{x-1}+x^x{\log }_e^x\left({\log }_e^{e}x\right)\right]$ .