Differential equation about Complex functions. $z\in \mathbb{C}$
$f(z)$ : meromorphic function in a connected domain
I cannot solve this differential equation.
\begin{equation}
f'(z)=f(z)^2 \cdots \ast
\end{equation}
My mathematic book says "Integrate the equation $\ast$ for $f$.
But I cannot know what I should do.
If I integrate $\ast$ for $f$,
\begin{equation}
f(z)=\int f(z)^2 dz
\end{equation}
I cannot proceed from this.
And I tried by another way.
I represent $f(z)=\sum_{n=0}^{\infty} c_n z^n$, and the equation $\ast$ is
\begin{equation}
\sum_{n=0}^{\infty} (n+1)c_{n+1} z^n
=\left(
\sum_{n=0}^{\infty} c_n z^n \right)^2
\end{equation}
But this approach doesn't seem to work.
So I want to solve by integrating the equation $\ast$ for $f$, as my book says.
What should I do?
 A: One might proceed as follows: in any region where
$f(z) \ne 0, \; \; (1) \tag 1$
we have
$\left (\dfrac{1}{f(z)} \right )' = (f^{-1}(z))' = -f^{-2}(z)f'(z) = -\dfrac{f'(z)}{f^2(z)} 
;  \; \; (2) \tag 2$
now if
$f'(z) = f^2(z), \tag 3$
then
$\dfrac{f'(z)}{f^2(z)} = 1, \tag 4$
so in accord with (2),
$\left (\dfrac{1}{f(z)} \right )' = -1; \tag 5$
we may integrate this along a path 'twixt $z_0$ and $z$:
$\dfrac{1}{f(z)} - \dfrac{1}{f(z_0)} = \displaystyle \int_{z_0}^z \left (\dfrac{1}{f(w)} \right )'\; dw = \int_{z_0}^z (-1)\; dw = z_0 - z; \tag 6$
we now solve for $f(z)$ algebraically:
$\dfrac{1}{f(z)} = z_0 - z + \dfrac{1}{f(z_0)} = \dfrac{f(z_0)(z_0 - z) + 1}{f(z_0)}, \tag 7$
or
$f(z) = \dfrac{f(z_0)}{f(z_0)(z_0 - z) + 1} = f(z_0)(f(z_0)(z_0 - z) + 1)^{-1}, \tag 8$
which may easily be checked by direct differentiation; from (8),
$f'(z) = -f(z_0)(f(z_0)(z_0 - z) + 1)^{-2}(-f(z_0)) = f^2(z_0)(f(z_0)(z_0 - z) + 1)^{-2} = f^2(z), \tag 9$
which is (3).  We also see that (8) yields
$f(z_0) = f(z_0)(f(z_0)(z_0 - z_0) + 1)^{-1} = f(z_0)(1^{-1}) = f(z_0), \tag{10}$
which demonstrates that $f(z)$ satisfies the appropriate boundary/initial condition at $z_0$.
It is perhaps worth observing that the hypothesis (1) is satisfied by (8) for all $z \in \Bbb C$ as long as $f(z_0) \ne 0$, that is, the solution is non-trivial.
