First pick one to be dependent on the other, for convenience and tradition's sake I'll say that #x# depends on #y#.
#9 = y(e^(2y))/x#
Differentiate both sides
#0 = d/dx(y(e^(2y))/x)#
Use the product rule
#0 = (e^(2y))/x*dy/dx + yd/dx(e^(2y))/x#
And again (or use the quotient rule)
#0 = (e^(2y))/x*dy/dx + y(1/x*d/dx(e^(2y)) + e^(2y)d/dx*1/x)#
#0 = (e^(2y))/x*dy/dx + y(1/x*d/dx(e^(2y)) -e^(2y)/x^2)#
Say that #2y = u#
#0 = (e^(2y))/x*dy/dx + y(1/x*d/(du)e^(u)*(du)/dx -e^(2y)/x^2)#
#0 = (e^(2y))/x*dy/dx + y(e^(2y)/x*(du)/dx -e^(2y)/x^2)#
Since #2y = u#, #(du)/dx = 2dy/dx#
#0 = (e^(2y))/x*dy/dx + 2ye^(2y)/x*dy/dx -ye^(2y)/x^2#
But since #9 = ye^(2y)/x#
#0 = 9/y*dy/dx + 18*dy/dx -9/x#
Put #dy/dx# in evidence
#9/x = dy/dx(9/y + 18)#
Pass it over dividing
#9/(x(9/y+18)) = dy/dx#
Use algebra to make it look prettier if you wish
#9/(9x/y+18x) = dy/dx#
#1/(x/y+2x) = dy/dx#
#y/(x+2xy) = dy/dx#
If you need #dx/dy# just take the inverse of #dy/dx#
