The necessary and sufficient conditions for the solution of the equation $\frac{dy}{dx} = f(y)$ is locally unique. $$\frac{\mathrm{d}y}{\mathrm{d}x} = f(y)$$
where $f(y)$ is continuous on $|y-a|\leq \epsilon$,and $f(y)=0$ iff $y=a$.
To Proof : For the initial value point on $y=a$,the equation has local unique solution iff $\left|\int_a^{a+\epsilon}\frac{\mathrm{d}y}{f(y)}\right|= \infty$ 
How to proof Initial value $\Rightarrow$ $\left|\int_a^{a+\epsilon}\frac{\mathrm{d}y}{f(y)}\right|= \infty$ ?
 A: It's not true.  Consider e.g. $$\dfrac{dy}{dx} = - y^{1/3}, y(0) = 0$$
Even though $\int_0^{\epsilon} - y^{-1/3}\ dy$ is finite, it's obvious that the only solution is $y = 0$. 
What is true is this.  Suppose $f(y) > 0$ for $a+\epsilon > y > a$ and $\displaystyle\int_{a}^{a+\epsilon} \dfrac{dy}{f(y)} = b < \infty$.  Then besides the constant solution $y = a$, there is a solution
defined implicitly by $\displaystyle\int_a^y \dfrac{ds}{f(s)} = x$ for $0 < x < b$, with $y = 0$ for $x \le 0$.
Similarly, if $f(y) < 0$ for $a-\epsilon < y < a$ and $\displaystyle \int_{a-\epsilon}^a \dfrac{dy}{f(y)} = c > -\infty$, there is another solution in $c < x < 0$. 
A: I don't quite understand the counterexample Robert Israel gives. This equation obviously has an infinite number of solutions.I'd like to give a proof, though it's too late for that.
${\rm{Proof:}}$
Assume that $(x_0,y_0)\in\left\{(x,y):0\leq|y-a|\leq\varepsilon\right\}$.
WLOG $y_0>a.$ By Peano's Theorem there must be a solution $y=\varphi(x)$ with initial value $y_0=\varphi(x_0).$
Thus if $x\in(-\infty,x_0),$ we have $\varphi(x)$ increase.(Because $\varphi^\prime(x)=f(\varphi(x))>0$). Since the solution of $\frac{{\operatorname{d}}y}{{\operatorname{d}}x}=f(y)$ is locally unique, $\varphi(x)>a,\quad x\in(-\infty ,x_0).$
Let$b=\lim_{x\to-\infty}\varphi(x)\Rightarrow b\geq a$, thus $$\int_b^{y_0}\frac{{\operatorname{d}}y}{f(y)}=\int_{-\infty}^{x_0}{\operatorname{d}}x=\infty\Rightarrow \left|\int_a^{a+\varepsilon}\frac{{\operatorname{d}}y}{f(y)}\right|=\infty.$$
