Double logarithmic differentiation

$\begin{aligned} y & = x^{\ln^x x} & \small \color{#3D99F6} \text{Taking natural logarithm both sides.} \\ \ln y & = \ln^x x \cdot \ln x \\ & = \ln^{x+1} x & \small \color{#3D99F6} \text{Taking natural logarithm both sides.} \\ \ln (\ln y) & = (x+1) \ln (\ln x) & \small \color{#3D99F6} \text{Differentiating both sides.} \\ \frac 1{\ln y} \cdot \frac 1y \cdot \frac {dy}{dx} & = \ln (\ln x) + (x+1)\cdot \frac 1{\ln x} \cdot \frac 1x \\ \implies \frac {dy}{dx} & = y \ln y \left(\ln (\ln x) + \frac {x+1}{x\ln x} \right) \end{aligned}$

$\implies f(x) = \boxed{\ln(\ln x) + \dfrac {x+1}{x\ln x}}$