Given implicit function #\frac{xy^2-3xy}{\sqrt{yx}-xy}=C#
Differentiating the above function wrt #x# on both the sides we get
#\frac{(\sqrt{yx}-xy)\frac{d}{dx}(xy^2-3xy)-(xy^2-3xy)\frac{d}{dx}(\sqrt{yx}-xy)}{(\sqrt{yx}-xy)^2}=0#
#(\sqrt{yx}-xy)(2xyy'+y^2-3xy'-3y)-(xy^2-3xy)(1/2\sqrt{x/y}y'+1/2\sqrt{y/x})=0\quad (\forall \ xy\ne0,1)#
#(2(\sqrt{xy}-xy)(2xy-3x)-\sqrt{x/y}(x^2y-3xy))y'=\sqrt{y/x}(x^2y-3xy)-2(\sqrt{xy}-xy)(y^2-3y)#
#y'=\frac{\sqrt{y/x}(x^2y-3xy)-2(\sqrt{xy}-xy)(y^2-3y)}{2(\sqrt{xy}-xy)(2xy-3x)-\sqrt{x/y}(x^2y-3xy)}#
Now, setting #x=2, y=5# in above equation, we get the slope #y'# of the tangent line
#y'=\frac{\sqrt{5/2}(2^2\cdot5-3(2)(5))-2(\sqrt{2\cdot 5}-2\cdot5)(5^2-3\cdot 5)}{2(\sqrt{2\cdot 5}-2\cdot 5)(2\cdot 2\cdot 5-3\cdot 2)-\sqrt{2/5}(2^2\cdot5-3\cdot 2\cdot 5)}#
#y'=120.9430585/-185.1316702#
#y'=-0.653281301#
