First, write it as #(xy)^{1/2}=x-2y# or #x^{1/2}y^{1/2}=x-2y#.
Next, differentiate both sides with respect to #x#, assuming that #y# is a function of #x#. You'll need the Product Rule and the Chain Rule:
#\frac{1}{2}x^{-1/2}y^{1/2}+\frac{1}{2}x^{1/2}y^{-1/2}\cdot dy/dx=1-2 dy/dx#.
Finally, solve this equation for #dy/dx#:
#dy/dx(\frac{1}{2}x^{1/2}y^{-1/2}+2)=1-\frac{1}{2}x^{-1/2}y^{1/2}#
#dy/dx=\frac{1-\frac{1}{2}x^{-1/2}y^{1/2}}{\frac{1}{2}x^{1/2}y^{-1/2}+2}=\frac{1-\frac{\sqrt{y}}{2\sqrt{x}}}{\frac{\sqrt{x}}{2\sqrt{y}}+2}=\frac{2\sqrt{xy}-y}{x+4\sqrt{xy}}#
