Differentiate both sides with respect to one variable, for ease's sake I'll differentiate with respect to #x#.
#d/dx6 = d/dx ln(y)/(e^x-e^y) - d/dxe^x - d/dxe^y#
Assume the other variable is dependant on the one you're differentiating and use the chain rule to be able to differentiate them
#0 = 1/(e^x - e^y)*d/dxln(y) - ln(y) d/dx1/(e^x-e^y) - e^x - d/dxe^y#
#0 = 1/(e^x - e^y)*1/y*dy/dx - ln(y) d/dx1/(e^x-e^y) - e^x - e^ydy/dx#
For the middle one, say #e^x - e^y = u# so #(du)/dx = e^x - e^ydy/dx#
#0 = 1/(e^x - e^y)*1/y*dy/dx + ln(y)/(e^x-e^y)^2*(e^x - e^ydy/dx) - e^x - e^ydy/dx#
Use algebra to isolate the derivate of the other variable
#0 = 1/(y(e^x-e^y))dy/dx +e^x ln(y)/(e^x-e^y)^2 -e^yln(y)/(e^x-e^y)^2
dy/dx - e^x -e^ydy/dx#
#e^x(1 - ln(y)/(e^x-e^y)^2) = dy/dx(1/(y(e^x-e^y)) + e^yln(y)/(e^x-e^y)^2 - e^y)#
#e^(x)/(e^x-e^y)^2((e^x-e^y)^2 - ln(y)) = dy/dx((e^x-e^y + ye^yln(y)-e^yy(e^x-e^y)^2)/(y(e^x-e^y)^2))#
#e^(x)((e^x-e^y)^2 - ln(y)) = dy/dx((e^x-e^y + ye^yln(y)-e^yy(e^x-e^y)^2)/(y))#
#dy/dx = ye^(x)((e^x-e^y)^2-ln(y))/(e^x-e^y+ye^yln(y)-ye^y(e^x-e^y)^2)#
From there it's just a matter of simplifying, if you took #dx/dy# and need #dy/dx# just flip the denominator and numerator.
Dividing by #(e^x-e^y)# in top and below
#dy/dx = ye^(x)((e^x-e^y)-ln(y)/(e^x-e^y))/(1+ye^yln(y)/(e^x-e^y)-ye^y(e^x-e^y))#
From the original statemement we know that
#6 = ln(y)/(e^x-e^y) - e^x -e^y#
#6 + e^y = ln(y)/(e^x-e^y) -e^x#
#-6-e^y = e^x -ln(y)/(e^x-e^y)# and #6+ e^y -e^x = ln(y)/(e^x-e^y)#
#dy/dx = ye^(x)(-6-2e^y)/(1+ye^y(6+e^y-e^x)-ye^y(e^x-e^y))#
#dy/dx = ye^(x)(-6-2e^y)/(1+6ye^y+ye^(2y)-ye^(x+y)-ye^(x+y)+ye^(2y))#
#dy/dx = ye^(x)(-6-2e^y)/(1+6ye^y+2ye^(2y)-2ye^(x+y))#
#dy/dx = -ye^(x)(6+2e^y)/(1+6ye^y+2ye^(2y)-2ye^(x+y))#
