Implicit Turning Point

$y=x^{x^2} \\ \ln{y}=x^2\ln{x} \\ \frac{1}{y}\frac{dy}{dx} = x(2\ln{x} + 1)\\ \frac{dy}{dx} = xy(2\ln{x} + 1) \\ \frac{dy}{dx} = x\times x^{x^2}(2\ln{x} + 1) \\ \frac{dy}{dx} = x^{x^2 + 1}(2\ln{x} + 1) = 0\\ x^{x^2 + 1} = \ce{0 -> no solutions in domain} \\ (2\ln{x} + 1) = 0 \\ x = e^{-1/2} \\ x = \frac{1}{\sqrt{e}}$

$\begin{aligned} y & = f(x) = x^{x^2} = e^{x^2\ln x} \\ \implies \frac {dy}{dx} & = (2x\ln x + x) x^{x^2} \end{aligned}$

Turning point occurs when $\dfrac {dy}{dx} = 0$ . That is:

$\begin{aligned} x (2\ln x + x) x^{x^2} & = 0 & \small \color{#3D99F6} \text{For }x > 0 \\ \implies 2\ln x + 1 & = 0 \\ \ln x & = - \frac 12 \\ x & = e^{-\frac 12} = \boxed{\frac 1{\sqrt e}} \end{aligned}$