If $y=x^{x^{x^{x^{x^{...}}}}}$, then $\frac{d y}{d x}$ is equal to

$\frac{y^2}{x(1-y \log x)}$

We have,

$y=x^{x^{x^{x^{x^{...}}}}}$

$\Rightarrow y=x^y$

$\Rightarrow y=e^{y \log x}$

$\Rightarrow \frac{d y}{d x}=e^{y \log x} \frac{d}{d x}(y \log x)$

$\Rightarrow \frac{d y}{d x}=x^y\left(\frac{d y}{d x} \log x+\frac{y}{x}\right)$

$\Rightarrow \frac{d y}{d x}=y\left(\frac{d y}{d x} \log x+\frac{y}{x}\right)$           $\left[∵ y=x^y\right]$

$\Rightarrow \frac{d y}{d x}=\frac{y^2}{x(1-y \log x)}$