Suppose $\dim L=3$ and $L=[L,L]$. Prove that $L$ must be simple. [Observe first that any homomorphic image of $L$ also equals its derived algebra.] Recover the simplicity of $\mathfrak{sl}(2,F)$, $\operatorname{char}F\ne 2$.
First, the part in brackets. Suppose the homomorphic image of $L$ under $\varphi$ is $M$. Since $M$ is an algebra, we must have $[M,M]\subseteq M$. Now, consider an element $\varphi(z)\in M$. Since $L=[L,L]$, we have that $$z=a_1[x_1,y_1]+\ldots+a_k[x_k,y_k]$$ for scalars $a_1,\ldots,a_k\in F$ and $x_1,\ldots,x_k,y_1,\ldots,y_k\in L$. Then $$\varphi(z)=a_1\varphi([x_1,y_1])+\ldots+a_k\varphi([x_k,y_k])=a_1[\varphi(x_1),\varphi(y_1)]+\ldots+a_k[\varphi(x_k),\varphi(y_k)],$$ showing that $M\subseteq [M,M]$.
Now suppose $\dim L=3$ and $L=[L,L]$, and that $L$ is not simple. So $L$ has an ideal $I$ that is neither $0$ nor $L$. So $I$ has dimension $1$ or $2$. What then?
If $I$ is a non-trivial ideal of $L$, then $M=L/I$ has dimension one or two. In two earlier questions you have shown that this implies that $[M,M]$ is a proper subspace of $M$. This violates your observation about homomorphic images of $L$.