If y = log_e$(\frac{2x}{1-x})$, then $\frac{d^2y}{dx^2}$ at $x = \frac{1}{2}$ is

0

$t=log_e(\frac{2x}{1-x})$ ⇒ So, $\frac{dy}{dx}=\frac{\frac{1}{2x}}{1-x}×\frac{d}{dx}(\frac{2x}{1-x})$

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

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

at $x=\frac{1}{2}$

$\frac{d^2y}{dx^2}=\frac{1}{(1-\frac{1}{2})^2}-\frac{1}{(\frac{1}{2})^2}⇒\frac{1}{(\frac{2-1}{2})^2}-\frac{1}{(\frac{1}{4})}$

$\frac{1}{(\frac{1}{2})^2}-\frac{1}{\frac{1}{4}}=\frac{1}{(\frac{1}{4})}-\frac{1}{(\frac{1}{4})}⇒4-4=0$

Correct option is 3rd.