I want to show that the Lie algebra $A=\{ x\in M_3(\mathbb{C}) : x^T\begin{pmatrix} 1 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 1\\ \end{pmatrix}+\begin{pmatrix} 1 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 1\\ \end{pmatrix}x=0\}$ is not simple.
I've tried to look at the center of $A$, since the center is an ideal. However it seems that the center is trivial (I haven't managed to find anything).
Using the given relation, then I've found out that elements of $A$ are on the form $x=\begin{pmatrix} a & b & c\\ d & 0 & d\\ -a & -b-d & -c\\ \end{pmatrix}$.
I've also tried to construct a non-trivial ideal of $A$, but without any luck.
Now I'm kinda stuck. Any hint is appriciated.
I assume that the field $K$ is $\Bbb C$ or $\Bbb R$. Then we know by the classification that there is no simple $4$-dimensional Lie algebra over $K$.
We can also see this explicitly in your case over any field $K$ of characteristic zero as follows. Let $(x_1,x_2,x_3,x_4)$ denote the basis, by setting $a,b,c,d$ equal to $1$ successively and the other parameters to zero. Then we can compute the explicit Lie brackets: $$ [x_1,x_2]=[x_2,x_3]=[x_4,x_3]=x_2, [x_1,x_3]=[x_2,x_4]=x_1+x_3. $$ This Lie algebra $L$ does not satisfy $[L,L]=L$, so it is not perfect and hence not simple. In fact, it is solvable, since the derived series is given by $[L,L]=\langle x_1+x_3,x_2\rangle$ and $[[L,L],[L,L]]=0$. Here we also see that the derived ideal $[L,L]$ is a proper ideal.