= Solution
The <orbital angular momentum> $L_3=X_1P_2-X_2P_1$ generates rotations in the plane. The <canonical commutation relations> give
$$
[L_3,X_1]=i\hbar X_2,\qquad [L_3,X_2]=-i\hbar X_1,\qquad
[L_3,P_1]=i\hbar P_2,\qquad [L_3,P_2]=-i\hbar P_1.
$$
Use the <commutator product rule> $[A,B^2]=[A,B]B+B[A,B]$. Since different coordinates commute,
$$
\begin{aligned}
[L_3,P_1^2+P_2^2]&=2i\hbar P_1P_2-2i\hbar P_2P_1=0,\\
[L_3,\omega_1^2X_1^2+\omega_2^2X_2^2]
&=2i\hbar(\omega_1^2-\omega_2^2)X_1X_2.
\end{aligned}
$$
Thus, for arbitrary positive <angular frequencies>,
$$
[L_3,H]=i\hbar m(\omega_1^2-\omega_2^2)X_1X_2.
$$
At $\omega_1=\omega_2$, this becomes $\boxed{[L_3,H]=0}$. The <rotational symmetry> of the <two-dimensional isotropic harmonic oscillator> allows a basis of simultaneous <energy eigenstates> and <orbital angular momentum> <eigenstates>.
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