Differentiating implicitly gets us
#2y+2xdy/dx = 2y dy/dx#
So that, #dy/dx = y/(y-x)#
Further discussion
#2xy=y^2# is equivalent to #y^2-2xy = 0# which in turn is equivalent to #y(y-2x) = 0#.
The graph of this equation consists of the line #y=0# (the #x# axis) and the line #y=2x#
graph{y(y-2x)=0 [-17.19, 14.86, -9.43, 6.59]}
Description of Cases
At #(0,0)# there is no derivative.
For points #(x,0)# with #x != 0#, we have #dy/dx = 0#.
Other points on the graph lie on #y=2x# alone and have derivative #dy/dx = 2#
Rewriting the derivative
Our result above says #dy/dx = y/(y-x)#.
For #y != 0#, we can write #dy/dx = y^2/(y^2-xy)#
Since #y^2 = 2xy# (see the original equation), we can write
#dy/dx = y^2/(2xy-xy) = y^2/(xy) = y/x#
Checking the three cases, we see that
At #(0,0)# we have #y/x# does not exist..
For points #(x,0)# with #x != 0#, we have #y/x = 0#.
Other points on the graph have #y=2x# alone and have derivative #y/x = (2x)/x = 2#.
Therefore, #dy/dx = y/x# is also true.
