When $\operatorname{char}F=0$, show that each classical algebra $L=A_l,B_l,C_l$ or $D_l$ is equal to $[LL]$.
Since $L$ is a Lie algebra, we know that if $x,y\in L$, then $[x,y]=xy-yx\in L$. So $[LL]\in L$. We have to show that $L\in [LL]$, i.e. every element $z$ of $L$ is equal to $xy-yx$ for some $x,y\in L$. How might we choose $x,y\in L$ so that $xy-yx=z$?
It should be helpful to know that $L$ is a simple Lie algebra, and that $[LL]$ is a Lie ideal in $L$. This means that there are only two possibilities; either $[LL]=L$ or $[LL]=0$. Why can we exclude the latter?
This answers your question, although it doesn't show how to compute $x$ and $y$ given $z$ as you've asked. Such a constructive proof may be much more difficult.