Calculus Question- Help!

Find the derivative of the following via implicit differentiation:
d/dx(e^x + y) = d/dx(y)

Differentiate the sum term by term:
d/dx(e^x) + d/dx(y) = d/dx(y)

Using the chain rule, d/dx(e^x) = ( d e^u)/( du) ( du)/( dx), where u = x and d/( du)(e^u) = e^u:
d/dx(y) + e^x (d/dx(x)) = d/dx(y)

The derivative of x is 1:
d/dx(y) + 1 e^x = d/dx(y)

Using the chain rule, d/dx(y) = ( dy(u))/( du) ( du)/( dx), where u = x and d/( du)(y(u)) = y'(u):
e^x + (d/dx(x)) y'(x) = d/dx(y)

The derivative of x is 1:
e^x + 1 y'(x) = d/dx(y)

Using the chain rule, d/dx(y) = ( dy(u))/( du) ( du)/( dx), where u = x and d/( du)(y(u)) = y'(u):
e^x + y'(x) = (d/dx(x)) y'(x)

The derivative of x is 1:
e^x + y'(x) = 1 y'(x)

Simplify the expression:
e^x + y'(x) = y'(x)

Subtract y'(x) from both sides:
e^x = 0

Subtract e^x from both sides:
0 = -e^x

Divide both sides by 0:
y'(x) = ∞

Definitely cannot divide by 0.     LOL

dy/dx for e^x + y = y

Perhaps I do not understand the question. Maybe more than just perhaps  

e^x + y = y

e^x=0

Never happens so you cannot find a tangent.   ???

e^x and also its gradient do approach 0 as x approaches -infinity though. 

Maybe another mathematician should look at this. Maybe my definitions are not quite right.

the question does not make sense to me.