#1=x/y-e^(xy)#
First we have to know that we can differentiate each part separately
Take #y=2x+3# we can differentiate #2x# and #3# seperately
#dy/dx=dy/dx2x+dy/dx3 rArrdy/dx=2+0#
So similarly we can differentiate #1#, #x/y# and #e^(xy)# separately
#dy/dx1=dy/dxx/y-dy/dxe^(xy)#
Rule 1: #dy/dxC rArr 0# derivative of a constant is 0
#0=dy/dxx/y-dy/dxe^(xy)#
#dy/dxx/y# we have to differentiate this using the quotient rule
Rule 2: #dy/dxu/v rArr ((du)/dxv-(dv)/dxu)/v^2# or #(vu'-uv')/v^2#
#u=x rArr u'=1#
Rule 2: #y^n rArr (ny^(n-1)dy/dx)#
#v=y rArr v'=dy/dx#
#(vu'+uv')/v^2=(1y-dy/dxx)/y^2#
#0=(1y-dy/dxx)/y^2-dy/dxe^(xy)#
Lastly we have to differentiate #e^(xy)# using a mixture of the chain and the product rule
Rule 3: #e^u rArr u'e^u#
So in this case #u=xy# which is a product
Rule 4: #dy/dxxy=y'x+x'y#
#x rArr 1#
#y rArr dy/dx#
#y'x+x'y=dy/dxx+y#
#u'e^u=(dy/dxx+y)e^(xy)#
#0=(1y-dy/dxx)/y^2-(dy/dxx+y)e^(xy)#
Expand out
#0=(1y-dy/dxx)/y^2-dy/dxxe^(xy)+ye^(xy)#
Times both sides by #y^2#
#0=y-dy/dxx-dy/dxxe^(xy)y^2+ye^(xy)y^2#
#0=y-dy/dxx-dy/dxxe^(xy)y^2+e^(xy)y^3#
Place all the #dy/dx# terms on one side
#y-e^(xy)y^3=dy/dxx-dy/dxxe^(xy)y^2#
Factorize out #dy/dx# on the RHS (right hand side)
#-y-e^(xy)y^3=dy/dx(x-xe^(xy)y^2)#
#(-(y+e^(xy)y^3))/(x-xe^(xy)y^2)=dy/dx#
