At which $x > 0$ is the derivative of $y = x^{1/x}$ zero?

Explanation

1. 01

   $\log y = \frac{\log x}{x}$, so $\frac{y'}{y} = \frac{1 - \log x}{x^{2}}$.

2. 02

   $y > 0$, so $y' = 0$ exactly when $\log x = 1$: $x = e$.