If, show that

The given relationship is $e^{y}(x+1)=1$

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

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

Taking logarithm on both the sides, we obtain

$y=\log \frac{1}{(x+1)}$

Differentiating this relationship with respect to x, we obtain

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

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

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

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

Hence, proved.