Clearly, $\displaystyle x^{-\sqrt{x}}\ne0$ so that $\displaystyle f'(x)=0$ precisely when $\displaystyle \frac{1}{\sqrt{x}}+\frac{\ln(x)}{2\sqrt{x}}=\frac{ 2+\ln(x)}{2\sqrt{x}}=0$ which is clearly true when $\displaystyle \ln(x)+2=0\implies \ln(x)=-2\implies x=\tfrac{1}{e^2}$.