If #y# and #x# are implicitly related ("depend on each other") as in the equation #xe^y-11x+5y=0#, it makes sense that the rate at which #y# changes (with respect to #x#) would depend on both #x# and #y#, and not just on #x#.
To find this rate of change #dy/dx#, what we need to do is take the derivative of both sides with respect to #x#, treating #y# as a function of #x#.
#" "xe^y" "-" "11x+" "5y=" "0#
#=>color(brown)(d/dx(xe^y))-color(blue)(d/dx11x)+color(green)(d/dx5y)=color(magenta)(d/dx0)#
#=>color(brown)((1)e^y+xe^y(dy/dx))-color(blue)11+color(green)(5(dy/dx))=color(magenta)0#
Notice: since #y# is considered a function of #x#, the derivative of #y# (with respect to #x#) remains its own term in the equation. To make reading the equation easier, we will use #y'# in place of #dy/dx#. Continuing, we have:
#e^y+xe^yy'-11+5y'=0#
#=>" "xe^yy'+5y'=11-e^y" "# (isolate the #y'# terms)
#=>" "y'(xe^y+5)=11-e^y" "# (factor out #y'#)
#=>" "y'=(11-e^y)/(xe^y+5)#
Therefore, the derivative of #y# with respect to #x# is
#dy/dx=(11-e^y)/(xe^y+5).#
