First we have to familrise ourselves with some calculs rules

#f(x)=2x+4# we can differentiate #2x# and #4# separately

#f'(x)=dy/dx2x+dy/dx4=2+0=2#

Similarly we can differentiate the #4#, #y# and #-(x-e^y)/(y-x)# separately

#dy/dx4=dy/dxy-dy/dx(x-e^y)/(y-x)#

We know that differentiating constants #dy/dx4=0#

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

Likewise the rule for differentiating y is #dy/dxy=dy/dx#

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

Lastly to differentiate #(x-e^y)/(y-x)# we have to use the quotient rule

Let #x-e^y=u#

and

Let #y-x=v#

The quotient rule is #(vu'-uv')/v^2#

#(du)/dx=(du)/dxx-(du)/dxe^y#

When deriving e we use the chain rule such that #e^y rArr (du)/dxe^y#

so #u'=1-dy/dxe^y#

#y-x=v#

so

#v'=(dv)/dxy-(dv)/dxx#

Using the same rules from above it becomes

#v'=dy/dx-1#

Now we have to do the quotient rule

#(vu'-uv')/v^2=((y-x)(1-(dy)/dxe^y)-(x-e^y)(dy/dx-1))/(y-x)^2#

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

Expand out

#0=dy/dx-((y-ydy/dxe^y-x+xdy/dxe^y)-(xdy/dx-x-e^ydy/dx+e^y))/(y-x)^2#

#0=dy/dx-(y-ydy/dxe^y-x+xdy/dxe^y-xdy/dx+x+e^ydy/dx-e^y)/(y-x)^2#

Multiply both sides by (#y-x)^2#

#0=dy/dx(y-x)^2-(y-ydy/dxe^y+xdy/dxe^y-xdy/dx+e^ydy/dx-e^y)#

#0=dy/dx(y-x)^2-y+ydy/dxe^y-xdy/dxe^y+xdy/dx-e^ydy/dx+e^y#

Place all the #dy/dx# terms on one side

#y-e^y=dy/dx(y-x)^2+ydy/dxe^y-xdy/dxe^y+xdy/dx-e^ydy/dx#

Factories dy/dx out of every term

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

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

#f'(x)=(y-e^y)/((y-x)^2+ye^y-xe^y+x-e^y)#