Has this I.V.P. unique solution?

With

$f(y) = \vert y \vert, \tag 1$

we have

$\vert f(y_1) - f(y_2) \vert = \vert \vert y_1 \vert - \vert y_2 \vert \vert \le \vert y_1 - y_2 \vert, \tag 2$

which shows that $f(y) = \vert y \vert$ is Lipschitz continuous with Lipschitz constant $1$; therefore by the Picard-Lindelof theorem, the equation

$\dot y = \vert y \vert \tag 3$

has a unique solution in some interval about any initial point $y_0 = y(t_0)$. So the only solution with $y(0) = 0$ is identically zero, $y(t) \equiv 0$.

This shouldn't really come as too much of a surprise since $\vert 0 \vert = 0$, so $0$ is an equilibrium of (3).

Finally, we don't need $f(y)$ differentiable, merely Lipschitz continuous, for existence and uniqueness to apply.

Note Added in Edit, Friday 17 August 2018 11:14 PM PST: To really flesh this answer out, we should probably also show that the unique solution $y(t) \equiv 0$ on an interval about $0$ extends uniquely to $(-\infty, \infty)$, but I'm leaving this discussion for a later date. End of Note.