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 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)#.
Further, as the question also involves product of variables, such as #-y^2x# and #-2xy#, we have to use Product Rule too. Product rule states if #f(x)=g(x)h(x)#
then #(df)/(dx)=(dg)/(dx)xxh(x)+(dh)/(dx)xxg(x)#
Hence using the two
#d/(dx)(-y^2x)=-y^2xxd/(dx) x-xd/(dx)y^2#
#=-y^2xx1-x xx 2yxx(dy)/(dx)=-y^2-2xy(dy)/(dx)# and
#d/(dx)(-2xy)=-2x xx (dy)/(dx)-2yxx d/(dx) x=-2x(dy)/(dx)-2y#
As #d/(dx)e^x=e^x#
implicitly differentiating #-1=-y^2x-2xy+e^x# we have
#0=-y^2-2xy(dy)/(dx)-2x(dy)/(dx)-2y+e^x# or
#2x(y+1)(dy)/(dx)=e^x-y^2-2y# or
#(dy)/(dx)=(e^x-y^2-2y)/(2x(y+1))#
