If (x+1) e^y=1, then :

$(x+1) e^y=1$

so $e^y=\frac{1}{1+x}$  ........(1)

differentiating both sides wrt x

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

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

from (1)

$\Rightarrow \frac{d y}{d x}=\frac{-1}{(1+x)}$

again differentiating wrt x

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

$\rightarrow \frac{d^2 y}{d x^2} =\frac{1}{(1+x)^2}$

$\frac{d^2 y}{d x^2} =\left(\frac{d y}{d x}\right)^2$