Generally differentiation problems involve functions i.e. #y=f(x)# - written explicitly as functions of #x#. However, some functions #y# are written implicitly as functions of #x#.
So 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)#.
One such equation is #y^2-xy=-5#.
Here differentiating #y^2# w.r.t. #y# gives #2y# and hence differential of #y^2# w.r.t. #x# is #2y(dy)/(dx)#. For #xy# we treat it as a product of #x# and #y# and use product formula. Differential of #-5# is of course #0# as it is constant. Hence differential of #y^2-xy=-5# is given by
#2y(dy)/(dx)-(1*y+x(dy)/(dx))=0#
or #(dy)/(dx)(2y-x)=y#
and #(dy)/(dx)=y/(2y-x)#
