Solving differential equation

I need help solving the DE below:

dz/dt + e^(t+z) = 0

I set it up as dz/dt = -e^(t+z) and then dz/dt = -e^z * e^t
After doing that I moved -e^z over by dividing
dz * -e^-z = dt * e^t

After integrating I get e^-z = e^t + C But I dont know where to go from here... I tried -t-c but to no avail

Help appreciated in advance.

As, you've done:

\(\displaystyle \L \int -e^{-z}dz = \int e^t dt \Longrightarrow e^{-z} = e^t + c\)

Take the natural log of both sides:

\(\displaystyle \L -z = t \Rightarrow z=-t + c\) (+ or - c doesn't matter)

Now, to see if this works: \(\displaystyle \frac{dz}{dt} = -1\), so:

\(\displaystyle \L \frac{dz}{dt} + e^{t+z} =\,\, -1 + e^{t+(-t)} \\ = -1+e^0 \\= -1+1 \\= 0\)
Yay!