We recall that, the slope of tgt. at pt. #(h,k)=[dy/dx]_((h.k))#
Hence, diff. both sides of the given eqn. of curve,
#d/dx[1+ln(xy)]=d/dx[e^(x-y)],#
#:. d/dx[1+lnx+lny]=e^(x-y)*d/dx(x-y)#.........[Chain Rule]
#:. 0+1/x+(1/y)dy/dx=e^(x-y){1-dy/dx}.#....[by Implicit diff. #d/dx(lny)=d/dylny*dy/dx#]
#:. 1/x+(1/y)dy/dx=e^(x-y)-e^(x-y)*dy/dx#
#:. (1/y)dy/dx+e^(x-y)*dy/dx=e^(x-y)-1/x#
#:.{(1+ye^(x-y))/y}dy/dx={xe^(x-y)-1}/x#
#:. dy/dx=(y(xe^(x-y)-1))/(x(1+ye^(x-y))#
#:.# The slope of tgt. at #(1,1)=[dy/dx]_(x=1,y=1)=0/2=0.#
#:.# The eqn. of tgt., using slop-point formula, is given by,
#y-1=0(x-1),# i.e., #y=1,# which is a horizontal line.
