If $y=x+e^x$, then $\frac{d^2 x}{d y^2}$ is equal to

$-\frac{e^x}{\left(1+e^x\right)^3}$

We have,

$y=x+e^x$

$\Rightarrow \frac{d y}{d x}=1+e^x$

$\Rightarrow \frac{d x}{d y}=\frac{1}{1+e^x}$

$\Rightarrow \frac{d^2 x}{d y^2}=\frac{d}{d y}\left(\frac{1}{1+e^x}\right)$

$\Rightarrow \frac{d^2 x}{d y^2}=-\frac{1}{\left(1+e^x\right)^2} \frac{d}{d y}\left(1+e^x\right)$

$\Rightarrow \frac{d^2 x}{d y^2}=-\frac{1}{\left(1+e^x\right)^2} e^x \frac{d x}{d y}=\frac{-e^x}{\left(1+e^x\right)^3}$