Answer
$${f_{xy}}\left( {x,y} \right) = {\text{ }}{f_{yx}}\left( {x,y} \right) = \frac{1}{{4\sqrt {xy} }}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \sqrt {xy} \cr
& f\left( {x,y} \right) = {x^{1/2}}{y^{1/2}} \cr
& {\text{Find the first partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right){\text{ then}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{x^{1/2}}{y^{1/2}}} \right] \cr
& {\text{treat }}y{\text{ as a constant}}{\text{, then}} \cr
& {f_x}\left( {x,y} \right) = {y^{1/2}}\frac{\partial }{{\partial x}}\left[ {{x^{1/2}}} \right] \cr
& {f_x}\left( {x,y} \right) = {y^{1/2}}\left( {\frac{1}{2}{x^{ - 1/2}}} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{1}{2}{x^{ - 1/2}}{y^{1/2}} \cr
& and \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^{1/2}}{y^{1/2}}} \right] \cr
& {\text{treat }}x{\text{ as a constant}}{\text{, then }} \cr
& {f_y}\left( {x,y} \right) = {x^{1/2}}\frac{\partial }{{\partial y}}\left[ {{y^{1/2}}} \right] \cr
& {f_y}\left( {x,y} \right) = {x^{1/2}}\left( {\frac{1}{2}{y^{ - 1/2}}} \right) \cr
& {f_y}\left( {x,y} \right) = \frac{1}{2}{x^{1/2}}{y^{ - 1/2}} \cr
& \cr
& {\text{Find the second partial derivatives }}{f_{xy}}\left( {x,y} \right){\text{ and }}{f_{yx}}\left( {x,y} \right){\text{ then}} \cr
& {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\frac{1}{2}{x^{ - 1/2}}{y^{1/2}}} \right] \cr
& {f_{xy}}\left( {x,y} \right) = \frac{1}{2}{x^{ - 1/2}}\frac{\partial }{{\partial y}}\left[ {{y^{1/2}}} \right] \cr
& {f_{xy}}\left( {x,y} \right) = \frac{1}{4}{x^{ - 1/2}}{y^{ - 1/2}} \cr
& {f_{xy}}\left( {x,y} \right) = \frac{1}{{4\sqrt {xy} }} \cr
& and \cr
& {\text{ }}{f_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\frac{1}{2}{x^{1/2}}{y^{ - 1/2}}} \right] \cr
& {\text{ }}{f_{yx}}\left( {x,y} \right) = \frac{1}{2}{y^{ - 1/2}}\frac{\partial }{{\partial x}}\left[ {{x^{1/2}}} \right] \cr
& {\text{ }}{f_{yx}}\left( {x,y} \right) = \frac{1}{{4\sqrt {xy} }} \cr
& \cr
& {\text{Then}}{\text{, we can verify that }}{f_{xy}}\left( {x,y} \right) = {\text{ }}{f_{yx}}\left( {x,y} \right) = \frac{1}{{4\sqrt {xy} }} \cr} $$