In implicit differentiation, one variable is not linked to another explicitly and it may not look like say #y=f(x)# but it will appear more as say #F(x,y)=0#.

The question is then how we evaluate #(dy)/(dx)#. We do this by differentiating both sides w.r.t #x# and wherever #y# appears, its differential is put as #(dy)/(dx)# and then evaluate #(dy)/(dx)#, which may be function of both #x# and #y#.

Let us do this using the given function #1=-xy+x+y#. Differentiating both sides we get

#0=-((dx)/(dx)*y+x(dy)/(dx))+1+(dy)/(dx)# (here we use product formula on #-xy#. So, we have

#0=-y-x(dy)/(dx)+1+(dy)/(dx)# or

#0=(1-x)(dy)/(dx)+1-y#

or #-(1-x)(dy)/(dx)=1-y# and hence

#(dy)/(dx)=-(1-y)/(1-x)#