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

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

Dividing both sides by $(x + 1)$:

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

Taking the natural logarithm of both sides:

$y = \ln\left(\frac{1}{x+1}\right)$

$y = -\ln(x+1)$

Differentiating with respect to $x$:

$\frac{dy}{dx} = -\frac{1}{x+1}$

Rewriting the first derivative as:

$\frac{dy}{dx} = -(x+1)^{-1}$

Differentiating again with respect to $x$:

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

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

Squaring the first derivative:

$\left(\frac{dy}{dx}\right)^2 = \left(-\frac{1}{x+1}\right)^2$

$\left(\frac{dy}{dx}\right)^2 = \frac{1}{(x+1)^2}$

Therefore:

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