If $e^{y}(x+1)=1,$ show that $\frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}\right)^{2}$.

Given the relationship $e^{y}(x+1)=1$.
Taking the natural logarithm on both sides:
$y + \ln(x+1) = \ln(1) = 0$
$y = -\ln(x+1)$
Differentiating with respect to $x$:
$\frac{dy}{dx} = -\frac{1}{x+1}$
Differentiating again with respect to $x$:
$\frac{d^{2}y}{dx^{2}} = -\left(-\frac{1}{(x+1)^{2}}\right) = \frac{1}{(x+1)^{2}}$
Since $\frac{dy}{dx} = -\frac{1}{x+1}$,then $\left(\frac{dy}{dx}\right)^{2} = \left(-\frac{1}{x+1}\right)^{2} = \frac{1}{(x+1)^{2}}$.
Thus,$\frac{d^{2}y}{dx^{2}} = \left(\frac{dy}{dx}\right)^{2}$.
Hence,proved.