Answer
$$\eqalign{
& {\text{Horizontal line at the point}}\left( {{e^{ - 2}},{e^{ - 2{e^{ - 1}}}}} \right) \cr
& {\text{The horizontal line is }}y = {e^{ - 2{e^{ - 1}}}} \cr} $$
Work Step by Step
$$\eqalign{
& y = {x^{\sqrt x }} \cr
& {\text{Use }}{b^x} = {e^{x\ln b}},{\text{ then}} \cr
& y = {e^{\sqrt x \ln x}} \cr
& {\text{Differentiating}} \cr
& \frac{{dy}}{{dx}} = {e^{\sqrt x \ln x}}\frac{d}{{dx}}\left[ {\sqrt x \ln x} \right] \cr
& \frac{{dy}}{{dx}} = {e^{\sqrt x \ln x}}\left[ {\sqrt x \left( {\frac{1}{x}} \right) + \frac{{\ln x}}{{2\sqrt x }}} \right] \cr
& \frac{{dy}}{{dx}} = {x^{\sqrt x }}\left[ {\frac{1}{{\sqrt x }} + \frac{{\ln x}}{{2\sqrt x }}} \right] \cr
& \frac{{dy}}{{dx}} = {x^{\sqrt x }}\left( {\frac{{2 + \ln x}}{{2\sqrt x }}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}{x^{\sqrt x - 1}}\left( {2 + \ln x} \right) \cr
& \cr
& {\text{Set the derivative equal to 0}} \cr
& \frac{1}{2}{x^{\sqrt x - 1}}\left( {2 + \ln x} \right) = 0 \cr
& {x^{\sqrt x - 1}} = 0{\text{ or }}2 + \ln x = 0 \cr
& {x^{\sqrt x - 1}}{\text{ is always positive}}{\text{, then}} \cr
& 2 + \ln x = 0 \cr
& \ln x = - 2 \cr
& x = {e^{ - 2}} \cr
& {\text{Then the function has a horizontal line at the point }} \cr
& \left( {\frac{1}{{{e^2}}},{{\left( {\frac{1}{{{e^2}}}} \right)}^{\sqrt {{e^{ - 2}}} }}} \right) \cr
& \left( {{e^{ - 2}},{e^{ - 2{e^{ - 1}}}}} \right) \cr
& {\text{The horizontal line is }}y = {e^{ - 2{e^{ - 1}}}} \cr} $$
