Given that #-3 = (xy)/(1-e^y)#, I assume the question asks for #dy/dx# to be expressed in terms of #x# and #y#, and that we are differentiating with respect to #x#. This works in principle if we differentiate with respect to #y#.

There are two methods to solve this.

Method 1: Brute force

#-3 = (xy)/(1-e^y)#

Differentiate with respect to #x# on both sides:

#0 = ((y+x dy/dx)(1-e^y)-(xy)(-e^y dy/dx))/(1-e^y)^2#

#0 = (y-ye^y+x dy/dx-xe^y dy/dx+xye^y dy/dx)/(1-e^y)^2#

#0 = (y-ye^y+dy/dx (x-xe^y+xye^y))/(1-e^y)^2#

#0 = y-ye^y+dy/dx (x-xe^y+xye^y)#

#xdy/dx (1-e^y+ye^y)=y(e^y-1)#

#dy/dx =(y(e^y-1))/(x(1-e^y+ye^y))# ... [1]

Method 2: Simplify before differentiating

#-3 = (xy)/(1-e^y)#

#3(e^y-1) = xy#

Differentiate with respect to #x# on both sides:

#3e^y dy/dx=y + x dy/dx#

#3e^y dy/dx - x dy/dx = y#

#dy/dx (3e^y - x) = y#

#dy/dx = y/(3e^y - x)# ... [2]

For additional practice, show that #x = (3(e^y-1))/y# and use this substitution to simplify both results, [1] and [2], into the expression #dy/dx = y^2/(3(1-e^y+ye^y))#