If $e^y(x+1)=1$. Then

$\frac{d^2y}{dx^2}=(\frac{dy}{dx})^3$ $(\frac{d^2y}{dx^2})^2=(\frac{dy}{dx})^3$ $\frac{dy}{dx}=\frac{d^2y}{dx^2}$ $\frac{d^2y}{dx^2}=(\frac{dy}{dx})^2$

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

We have $e^y=\frac{1}{x+1}$. Taking log in both sides we get that $y=\log (1/x+1)$. Differentiating both sides w.r.to x we get $\frac{dy}{dx}=(x+1)*{-1/(x+1)^2}=-1/x+1$. Differentiating again w.r.to x we get $\frac{d^2y}{dx^2}=1/(x+1)^2=(\frac{dy}{dx})^2$