Locate and classify the critical point using the Second Derivative Test

Using the quotient rule, $f'(x) = \dfrac{1 - \ln(x)}{x^2}$. Setting the numerator equal to zero gives $1 - \ln(x) = 0$, so $\ln(x) = 1$, which means $x = e$ is the only critical number for $x>0$ (the denominator $x^2$ is never zero or undefined on this domain).

Differentiating $f'(x) = \dfrac{1-\ln(x)}{x^2}$ again with the quotient rule gives $f''(x) = \dfrac{-3 + 2\ln(x)}{x^3}$. Evaluating at $x = e$: $f''(e) = \dfrac{-3 + 2(1)}{e^3} = \dfrac{-1}{e^3} < 0$. Since $f''(e) < 0$, the Second Derivative Test confirms $x = e$ is a local maximum, with value $f(e) = \dfrac{1}{e}$.

Setting $f''(x) = \dfrac{-3+2\ln(x)}{x^3} = 0$ requires the numerator to vanish (the denominator $x^3$ is never zero for $x>0$), so $2\ln(x) = 3$, giving $x = e^{3/2}$.

For $x$ slightly less than $e^{3/2}$, $\ln(x) < 3/2$, so $f''(x) < 0$ (concave down); for $x$ slightly greater, $\ln(x) > 3/2$, so $f''(x) > 0$ (concave up). Since concavity changes sign, $x = e^{3/2}$ is indeed an inflection point.