Is Lipschitz condition necessary for existence and uniqueness of a solution for a first order differential equation? Is the Lipschitz condition only a sufficient condition for existence and uniqueness of $\frac{dy}{dx}=f(x,y), y(0)=y_0$? Is it also necessary?
 A: No, it is necessary for local uniqueness. Existence can be covered with weaker conditions, see here or for even weaker conditions, here
A: First, it's worth pointing out that for a Cauchy problem
$$
\begin{cases}
y'=f(x,y) \\
y(x_0)=y_0
\end{cases}
$$
we require  $f$ to be a Lipschitz function only in the second variable, while it is enough for it to be continuous in the first one (this result is known as Picard-Lindelöf theorem).
A typical example for which uniqueness is lost is given by $f(x,y)=\sqrt{y}$ (which is indeed not Lipschitz).
As already pointed out, Peano's existence theorem (together with other more general results) guarantees existence if $f$ is just continuous in both variables.
Long story short: for a general ODE of the form $y'=f(x,y)$ being Lipschitz in the second variable is necessary for uniqueness, but not for existence.
A: Take the case of separable equations:
$$\frac{dy}{dx} = a(x) b(y)$$
with $x(x_0) = y_0$. Asume that $a>0$ around $x_0$. The equation is equivalent to
$$\int_{x_0}^x \frac{1}{a(t)} d t = \int_{y_0}^y b(t) dt$$
as a functional relationship between $x$ and $y$. For the existence and uniqueness of this type of Caychy problem near $(x_0, y_0)$ it is enough that $a>0$ and continuous, and $b$ is continuous. So Lipschitz in $y$ is not required. 
A: No, it is NOT necessary - It is only sufficient.
For example, if $f$ is JUST continuous and $f(x)>0$, for all $x\in\mathbb R$, then 
$$
x'=f(x), \quad x(\tau)=\xi, \tag{1}
$$
enjoys GLOBAL uniqueness, for all $(\tau,\xi)\in\mathbb R^2$.
Say $\varphi,\psi : I \to\mathbb R$, satisfy $(1)$, where $I$ is an open interval and $\tau\in I$. Then
$$
1=\frac{\varphi'(t)}{f\big(\varphi(t)\big)}=\frac{\psi'(t)}{f\big(\psi(t)\big)},
\quad \text{for all $t\in I$}.
$$
Set 
$$
F(x)=\int_\xi^x\frac{dx}{f(x)}, \quad x\in\mathbb R.
$$
Clearly, $F$ is continuously differentiable and strictly increasing (hence 1-1), in $\mathbb R$, and
$$
F\big(\varphi(t)\big)=\int_\tau^t\frac{\varphi'(t)}{f\big(\varphi(t)\big)}=\int_\tau^t\frac{\psi'(t)}{f\big(\psi(t)\big)}=F\big(\psi(t)\big), \quad\text{for all $t\in I$}.
$$
Thus, as $F$ is 1-1, $\varphi(t)=\psi(t)$, for all $t\in I$.
So, $(1)$ enjoys global uniqueness, while $f$ does not satisfy Lipschitz condition! 
