$\frac{dy}{dt} = (1 − y)(y + 3)$ Find equilibrium states of the differential equation below

$\frac{dy}{dt} = (1 − y)(y + 3)$

and describe whether they are stable or not

Ok to find equilibria we have to solve the equation: dN/dt = 0? right?

And then what?
We are asked to find the equilibrium points of 

$$\frac{dy}{dt} = (1 − y)(y + 3)$$

To find the equilibrium points, we set $\dfrac{dy}{dt} = 0$ and find the critical points, so we have

$$\frac{dy}{dt} = (1 − y)(y + 3) = 0 \implies y = 1, -3$$

Can you now figure out the stability of those two critical points?

A direction field plot shows:

$~~~~~~~~~~$ 

We can also draw a phase line as (compare to direction field plot):
This is a tough question, we are looking for solutions of y such that $\frac{dy}{dt} = 0$.

Now $ (1−y)(y+3) = y-y^2 +3-3y = -y^2 -2y +3.$ Note that $y=0$ implies $\frac{dy}{dt} = 3$ but when $y=2$ we have $\frac{dy}{dt} = -5$ and so IVT says there is a solution between y=0 and y=2. Unfortunately I can't narrow it down any more.