Let $T:M_{n\times n}(\mathbb{C})\to M_{n\times n}(\mathbb{C})$ be defined by $T(X)=A^2X+AXA+XA^2$ for some $A\in M_{n\times n}(\mathbb{C})$. Let $T:M_{n\times n}(\mathbb{C})\to M_{n\times n}(\mathbb{C})$ be defined by $T(X)=A^2X+AXA+XA^2$ for some $A\in M_{n\times n}(\mathbb{C})$. Then $T$ is a linear operator. The problem is that can we explicitly obtain the kernel of this linear map. If $A$ is a singular matrix then $X=xy^*$ satisfies $T(X)=0$ where $Ax=0$ and $y^*A=0$. But there exists matrix $A$ for which there is no connection between $A$ and $X$ for example if $$A=\begin{pmatrix}
1&0&0\\0&0&0\\0&1&0
\end{pmatrix} \qquad \text{and} \qquad X=\begin{pmatrix}
    0&0&0\\0&1&0\\0&2&1
\end{pmatrix}$$
So how we can obtain the kernel of $T$ completely.

Kronecker product is the right tool for analysing such issues. Here is how.
Let us begin by writing
$$T(X)=T_A(X)=A^2X+AXA+XA^2\tag{1}$$
in a different way, using the "vec" operator stacking the columns of any $3 \times 3$ matrix  into a single $9 \times 1$ column vector :
$$T(\operatorname{vec}(X))=\operatorname{vec}(A^2XI)+\operatorname{vec}(AXA)+\operatorname{vec}(IXA^2) \tag{3}$$
Now, if you are already familiar with Kronecker product, let us recall the fundamental identity :
$$\operatorname{vec}(AXB)=(B^T \otimes A) \operatorname{vec}(X)\tag{4}$$
Using (4) and denoting $\mathbf{X}=\operatorname{vec}(X)$, (3) can be transformed into  :
$$T(\mathbf{X})=(I \otimes (A^2))\mathbf{X}+(A^T \otimes A)\mathbf{X}+((A^2)^T \otimes I)\mathbf{X}$$
Otherwise said :
$$T(\mathbf{X})=\underbrace{((I \otimes (A^2))+(A^T \otimes A)+((A^2)^T \otimes I))}_{\mathbf{T}}\mathbf{X}\tag{5}$$
Otherwise said, operator $T$ is completely described by $9 \times 9$ matrix $\mathbf{T}$.
For example, if
$$A=\begin{pmatrix}
0&0&1\\1&0&0\\0&1&0
\end{pmatrix} \ \text{giving} \ A^2=\begin{pmatrix}
0&1&0\\0&0&1\\1&0&0
\end{pmatrix}$$
$\mathbf{T}$ is the sum of the following $9 \times 9$ matrices:
$$\mathbf{T}=\begin{pmatrix}
A^2&0&0\\0&A^2&0\\0&0&A^2
\end{pmatrix}+\begin{pmatrix}
0&A&0\\0&0&A\\A&0&0
\end{pmatrix}+\begin{pmatrix}
0&0&I\\I&0&0\\0&I&0
\end{pmatrix}=\begin{pmatrix}
A^2&A&I\\I&A^2&A\\A&I&A^2
\end{pmatrix}\tag{6}$$
How is all this connected with the initial question, obtaining the kernel of operator $T$ defined by (1) ?
Plainly, by finding the kernel of $\mathbf{T}$, then doing the inverse operation of $vec$, i.e., by "reshaping" all $9 \times 1$ column vectors constituting a basis of this kernel into $3 \times 3$ matrices.
Remark : In initial expression (1), $T(X)$ can be considered as the differential computed in $A$ of function $C$ defined by $C(X)=X^3$. I am almost sure that properties of operator $T$ can be deduced from this fact.
Edit : Having done extensive numerical simulations (using Matlab), I have found pairs :
$$(A,X) \ \text{solutions of eq. } T_A(X)=0$$
(see equ. (1)) where both $A$ and $X$ have full rank.
Here is a particularily interesting instance (see explanation below) :
$$A=\begin{pmatrix}
-1&1&-2\\2&0& \ \ 1\\2&1& \ \ 1
\end{pmatrix}, \ \ X=\begin{pmatrix}
-1&0&-1\\1&1&1\\1&1&0
\end{pmatrix}$$
Please note that $A$ is specific because $A^3=-3I_3$...
A remarkable fact is that for this choice of $A$, the kernel of $9 \times 9$ matrix $\mathbf{T}$ has dimension $6$ ! Moreover, we can exhibit a basis with $6$ elements of this kernel. Indeed, instead of matrix $X=X_1$ above, we could have taken one of the other five matrices :
$$X_2=\begin{pmatrix}
1&2&1\\1&1&-2\\1&0&-2
\end{pmatrix} \ \text{or} \ X_3=\begin{pmatrix}
-2&-1&1\\1&2&-1\\2&2&0
\end{pmatrix}  \ \text{or} $$
$$X_4=\begin{pmatrix}
0&0&-1\\1&1&1\\0&1&-1
\end{pmatrix}   \ \text{or} \ X_5=\begin{pmatrix}
2&1&0\\-2&-1&1\\0&2&-1
\end{pmatrix}   \ \text{or}$$
$$X_6=\begin{pmatrix}
0&-2&1\\-1&-1&1\\0&1&1
\end{pmatrix}, $$
One can check that all the $X_k, \  k=1,2 \cdots 6$ are full-rank matrices and that they constitute a basis of the kernel of $T_A$.
Of course, one can get, by combining these $X_k$, other matrices $X$ belonging to the kernel of $T£ which are rank-2 or even rank-1.
