If $y =\frac{log\, x}{x^2},$ then $\frac{d^2y}{dx^2}$ is equal to

$\frac{6logx-5}{x^4}$

Given

$y=\frac{\log x}{x^2}=(\log x)x^{-2}$

First derivative

$\frac{dy}{dx}=\frac{1}{x}x^{-2}+(\log x)(-2x^{-3})$

$=x^{-3}-2(\log x)x^{-3}$

$=\frac{1-2\log x}{x^3}$

Second derivative

$\frac{d^2y}{dx^2}=\frac{d}{dx}\left[\frac{1-2\log x}{x^3}\right]$

$=(1-2\log x)(-3x^{-4})+x^{-3}\left(-\frac{2}{x}\right)$

$=\frac{-3(1-2\log x)}{x^4}-\frac{2}{x^4}$

$=\frac{-3+6\log x-2}{x^4}$

$=\frac{6\log x-5}{x^4}$

$\frac{d^2y}{dx^2}=\frac{6\log x-5}{x^4}$