Show that

Here,

$f(x)=\frac{\log x}{x}$

$\Rightarrow f^{\prime}(x)=\frac{1-\log x}{x^{2}}$

For the local maxima or minima, we must have

$f^{\prime}(x)=0$

$\Rightarrow \frac{1-\log x}{x^{2}}=0$

$\Rightarrow 1=\log x$

$\Rightarrow \log e=\log x$

$\Rightarrow x=e$

Now,

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

$\Rightarrow f^{\prime \prime}(e)=\frac{-3+2 \log e}{e^{3}}=\frac{-1}{e^{3}}<0$

So, $x=e$ is the point of local maximum.