Implicit differentiation is a special case of the chain rule for derivatives. Generally differentiation problems involve functions i.e. #y=f(x)# - written explicitly as functions of #x#.
However, some functions like the one given above, are written implicitly as functions of #x# and #y#. Here, what we do is to treat #y# as #y=y(x)# and use chain rule. This means differentiating #y# w.r.t. #y#, but as we have to derive w.r.t. #x#, as per chain rule, we multiply it by #(dy)/(dx)#.
Thus as #x+y-1=ln(x^2+y^2)#, differentiating w.r.t. #x#, we have
#1+(dy)/(dx)=1/(x^2+y^2)(2x+2y(dy)/(dx))#
or #1-(2x)/(x^2+y^2)=(2y)/(x^2+y^2)(dy)/(dx)-(dy)/(dx)#
or #1-(2x)/(x^2+y^2)=[(2y)/(x^2+y^2)-1] (dy)/(dx)#
or #(x^2+y^2-2x)/(x^2+y^2)=[(2y-x^2-y^2)/(x^2+y^2)] (dy)/(dx)#
or #(dy)/(dx)=(x^2+y^2-2x)/(2y-x^2-y^2)#
