I assume that you want to find #d/dx(e^(x+y)+e^(2x)+e^(2y))#.

If I have misunderstood your question, please accept my apology.

#d/dx(e^(x+y)+e^(2x)+e^(2y)) = d/dx(e^(x+y))+d/dx(e^(2x))+d/dx(e^(2y))#.

For each term, we will need the chain rule applied to the exponential function:

#d/dx(e^u) = e^u (du)/dx#

#d/dx(e^(x+y)+e^(2x)+e^(2y)) = e^(x+y)d/dx(e^(x+y))+e^(2x)d/dx(2x)+e^(2y)d/dx(2y)#.

# = e^(x+y)(1+dy/dx) +2e^(2x)+e^(2y)*2dy/dx#

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

So,

#d/dx(e^(x+y)+e^(2x)+e^(2y))=e^(x+y)+e^(x+y)dy/dx +2e^(2x)+2e^(2y)dy/dx#

If we had an equation, we could solve for #dy/dx#, but as it is, all we can do is rewrite using algebra:

#d/dx(e^(x+y)+e^(2x)+e^(2y))=[e^(x+y)+2e^(2x)]+[e^(x+y) +2e^(2y)]dy/dx#