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)#. So for given function we proceed as under:

As #(xy)^2+xcos(xy)=ln(x/y)+ytan(xy)#, differentiating it w.r.t. #x# we have

#2xy(y+x(dy)/(dx))+cos(xy)+x(-sin(xy))(y+x(dy)/(dx))=1/(x/y)((y-x(dy)/(dx))/y^2)+tan(xy)(dy)/(dx)+ysec^2(xy)(y+x(dy)/(dx))#

or #2xy^2+2x^2y(dy)/(dx)+cos(xy)-xysin(xy)-x^2sin(xy)(dy)/(dx)=1/x-1/y(dy)/(dx)+tan(xy)(dy)/(dx)+y^2sec^2(xy)+xysec^2(xy)(dy)/(dx)#

= #(dy)/(dx)(2x^2y-x^2sin(xy)+1/y-tan(xy))=-2xy^2-cos(xy)+xysin(xy)+1/x+y^2sec^2(xy)#

and #(dy)/(dx)=(xysin(xy)+1/x+y^2sec^2(xy)-2xy^2-cos(xy))/(2x^2y-x^2sin(xy)+1/y-tan(xy))#

= #(x^2y^2sin(xy)+y+xy^3sec^2(xy)-2x^2y^3-xycos(xy))/(2x^3y^2-x^3ysin(xy)+x-xytan(xy))#