Consider the equation $\ddot s = s-s^3.$ Let $m=1.$
1) Write this as a first order system.
Let $\dot s=v.$ Then we get $\dot v=s-s^3.$ So first order system is $$\begin{pmatrix} \dot{s} \\ \dot{v} \end{pmatrix} = \begin{pmatrix} v \\ s - s^3 \end{pmatrix}$$
2) Compute the potential energy.
$$F=ma=\ddot s=s-s^3=\frac{-dV}{ds}=\frac{-d}{ds}\bigg(\frac{s^4}{4}-\frac{s^2}{2}\bigg).$$
So $V=\frac{s^4}{4}-\frac{s^2}{2}.$
3) Find all equilibria and classify them.
How do I do part 3)?
I'm moving my comments here cause I don't want to make the comment section too long.
An equilibrium is a constant solution to your differential equation. Since $s(t)$ is constant, it follows that its first derivative is zero. As you figured out, your equation has three possible equilibria: $s_1=-1,s_2=0,s_3=1$.
To "classify" an equilibrium means to establish whether it is stable or not. If you don't know what stable means, probably you need to review your notes or the textbook. Roughly speaking, an equilibrium $s_i$ is stable if, given an initial condition $s(0)=s_0$ that is "close" to $s_i$, the corresponding solution $s(t)$ will be "close" to $s_i$ for all times. If this does not happen, we say that the equilibrium is unstable. If, in addition to stay close to $s_i$, the solution is such that $\lim_{t\to\infty} s(t)=s_i$ we call the equilibrium asymptotically stable.
There are many ways to classify equilibria. If the system admits a potential, everything is easier though. In particular, an equilibrium $s_i$ is asymptotically stable if the potential has a local minimum at $s_i$, while it is unstable if the potential has a maximum at $s_i$.
I suggest you to go and review the theory about ODE, equilibria and stability, if you want to be able to tackle problems like this.