Derivative of (ln x)^ln x

Let $y=f(x)$ . Then $\ln{y}=\ln{x} \ln{\ln{x}}$ by logarithm rules. Differentiating implicitly, using product rule and chain rule, we get $\frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{x} \cdot \ln{\ln{x}} +\ln{x} \cdot \frac{1}{\ln{x}} \cdot \frac{1}{x}$ We know that $y= f(e) = 1^1 =1$ so plugging everything in we get $\frac{1}{1} \cdot \frac{dy}{dx}=0+1 \cdot 1 \cdot \frac{1}{e} \longrightarrow \frac{dy}{dx} = \boxed{\frac{1}{e}}$