Let $u(x,y)=e^{-y} \cos x$. Find all functions $v(x,y)$ such that $f(x+iy) = u(x, y) + iv(x, y)$ is complex differentiable for all $z ∈ \mathbb C$. 
Let $u(x,y) = e^{-y} \cos x$. Find all functions $v(x,y)$ such that $f(x+iy) = u(x, y) + iv(x, y)$ is complex differentiable for all $ z ∈ \mathbb C$.

I did this:
$Re (e^{-y}cosx + iv(x,y)) = e^{-y}cosx$

$u_x = -e^{-y} sinx = \frac{\delta}{\delta{y}} v$

$u_y = -e^{-y}cosx = - \frac{\delta}{\delta{x}} v$

$\implies v = \int (-e^{-y} sinx ) \delta{y} = e^{-y} sinx +f(x) $

$v=  \int (e^{-y} cosx ) \delta{x} = e^{-y} sinx +g(y) $
hece $v(x,y)= e^{-y} sinx$
is my solution correct?
and to add $f(z) = e^{-y} cosx = i e^{-y}sinx = e^{-y}.e^{ix} = e^{ix-y} $
am i right?

You're basically correct, with a minor glitch. I'd organize the exercise in a different way. The Cauchy-Riemann equations are
$$
\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}
\qquad
\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}
$$
Since
$$
\frac{\partial u}{\partial x}=-e^{-y}\sin x
\qquad
\frac{\partial u}{\partial y}=e^{-y}\cos x
$$
we need to solve
$$
\begin{cases}
\dfrac{\partial v}{\partial x}=-e^{-y}\cos x \\[6px]
\dfrac{\partial v}{\partial y}=-e^{-y}\sin x
\end{cases}
$$
Integrating the first equation with respect to $x$ yields
$$
v(x,y)=e^{-y}\sin x+\varphi(y)
$$
and differentiating with respect to $y$ yields
$$
-e^{-y}\sin x+\varphi'(y)=-e^{-y}\sin x
$$
so we conclude that $\varphi$ is a (real) constant, say $C$.
Hence
$$
v(x,y)=e^{-y}\sin x+C
$$
Your function is therefore
\begin{align}
f(x+iy)
&=e^{-y}\cos x+ie^{-y}\sin x+iC \\[6px]
&=e^{-y}(\cos x+i\sin x)+iC \\[6px]
&=e^{-y}e^{ix}+c \\[6px]
&=e^{ix-y}+iC
\end{align}
If $z=x+iy$, then $ix-y=iz$ and the function is
$$
f(z)=e^{iz}+iC
$$
