Given is that $Y = g(X)$ where $$g(X) = \left\{ X : X \geq 0\ \text{and } 0\ \text{for } X<0 \right\}$$
I understand that, since $g(X)$ is not strictly monotonic, we can't use the formula
$ f_y(y)= f_x(g^{-1}(y))\,|J(x,y)|$
So is was thinking about using
$F_y(y) = P(Y \leq y) = P(g(X) \leq y) = \{P(X\leq y) \text{ for } X\geq 0 \text{ and } P(0\leq y) \text{ for } X<0\} $
Now I don't really know how to proceed with this probability $P(0\leq y)$ to get $f_y(y)$, which can be computed by taking the derivative of $F_y(y)$
Not quite. The partition should be over the value of $y$ — because it is the argument of the CDF.
We have: $$Y=\begin{cases}~0&:& X < 0\\X&:& 0\leqslant X\end{cases}$$
So clearly $Y$ cannot be less than $0$, and if $y$ is at least $0$, then the event of $Y\,{\leqslant}\,y$ occurs exactly when $X\,{\leqslant}\,y$ occurs. Thus:
$$F_Y(y) = \begin{cases}~0&:& y<0\\F_X(y)&:& 0\leqslant y\end{cases}$$
Now if $X$ can be negative, then $F_X(0)\,{>}\,0$; which means there will be a step discontinuity at that point.