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

The given expression is $x=e^{y+e^{y+...}}$.
This can be written as $x=e^{y+x}$.
Taking natural logarithm on both sides:
$\ln x=y+x$.
$y=\ln x-x$.
Differentiating with respect to $x$:
$\frac{dy}{dx}=\frac{1}{x}-1=\frac{1-x}{x}$.