 Solution

Write $h=(h_1,h_2)$ for the homogeneous cubic correction and $A(x,y{)}^T=(y,0{)}^T$ for the nilpotent linear part. The chain rule gives the transformed cubic vector field as $f_3+Ah-DhA(x,y{)}^T$. Thus the cubic coefficients in the first equation arewhile those in the second equation areConsequently one explicit near-identity transformation isThe transformed equations have no cubic correction in $\dot{x}$, and only $x^3,x^{2}y$ survive in $\dot{y}$. The cubic coefficients areThis is the cubic elimination at a nilpotent double-zero point; the extra $3a_1$ is the nilpotent cubic damping invariant contribution from the first equation. Higher-order terms generated by the substitution are outside the cubic truncation.