The necessary and sufficient conditions for the solution of the equation $\frac{dy}{dx} = f(y)$ is locally unique.

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 $c < x < 0$.