Lie algebra is associative iff the derived subalgebra is contained in the centre 
Prove that a Lie algebra $\mathfrak g$ is associative iff the derived subalgebra of $\mathfrak g$ is contained in the centre of $\mathfrak g$, that is $\mathfrak g^{(1)} \subset c(\mathfrak g)$.

So we have the derived sub algebra is in the centre of g. We need to show that $(uv)w=u(vw)$. 
So take $u,v,w \in g$. Since $[u,[v,w]]=0$. I was thinking that you need to use jacobian identity on that. However, I'm stuck on what to do next. 
 A: You have one direction nearly finished:
The Jacobi identity says that $[u,[v,w]]+[v,[w,u]]+[w,[u,v]]=0$ thus $[u,[v,w]]=-[w,[u,v]]-[v,[w,u]]$. By skew-symmetry and linearity, $[u,[v,w]]=[[u,v],w]+[v,-[w,u]]$ then skew symmetry again $[u,[v,w]]=[[u,v],w]+[v,[u,w]]$. This is the derivation form of the Jacobi identity (which is usually more useful when proving things or thinking about properties). Notice this looks like the product rule for derivatives: $\frac{d}{dx}\left[vw\right] = \frac{dv}{dx}w+v\frac{dw}{dx}$.
Now your proposition follows since $[u,w]$ is in the derived subalgebra and thus also the center, $[v,[u,w]]=0$. Therefore, $[u,[v,w]]=[[u,v],w]$. Thus $\mathfrak{g}$ is associative.
For the other direction, suppose that $\mathfrak{g}$ is associative. Let $\sum\limits_{i=1}^n [a_i,b_i] \in \mathfrak{g}^{(1)}$. Then for $v\in\mathfrak{g}$,
By linearity and the derivation form of the Jacobi identity,
$$ \left[v,\sum\limits_{i=1}^n [a_i,b_i]\right] = \sum\limits_{i=1}^n [v,[a_i,b_i]] = \sum\limits_{i=1}^n \left([[v,a_i],b_i]+[a_i,[v,b_i]] \right)$$
Now use associativity, skew-symmetry, and linearity
$$ = \sum\limits_{i=1}^n \left([[v,a_i],b_i]+[[a_i,v],b_i] \right)
   = \sum\limits_{i=1}^n \left([[v,a_i],b_i]+[-[v,a_i],b_i] \right)
   = \sum\limits_{i=1}^n \left([[v,a_i],b_i]-[[v,a_i],b_i] \right) = 0$$
Thus $\sum\limits_{i=1}^n [a_i,b_i] \in \mathfrak{z}(\mathfrak{g})$ (the center of $\mathfrak{g}$).
