If $\mathrm{x}={\mathrm{e}}^{\mathrm{y}+{\mathrm{e}}^{\mathrm{y}+\dots \infty }}, \mathrm{x}>0,$ then $\frac{\mathrm{dy}}{\mathrm{dx}}$ is

Given $x=e^{y+e^{y+...}}$

This can be written as $x=e^{y+x}$

Apply logarithm on both sides and then differentiate

$\log x=\left(x+y\right)$

$\frac{1}{x}=1+\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\boxed{\frac{\mathbf{1}\mathbf{-}\mathbf{x}}{\mathbf{x}}}$