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I think similar questions always translate in the $L^p$ boundedness of a Fourier multiplier. In this case you want a Fourier multiplier which "exchanges the operator $D_1D_2D_3$ with the operator $D_1^3+D_2^3+D_3^3$. This opeartor (as seen by Plancherel) is (up to fixed constant) the Fourier multiplier by the function \begin{equation} m(\xi_1,\xi_2,\xi_3) = \frac{\xi_1\xi_2\xi_3}{\xi_1^3+\xi_2^3+\xi_3^3}. \end{equation} Then as Grafakos, Modern Fourier Analysis, Thrid Edition, Example 6.2.6 notices, when the multiplier has this typ of homogeneity, i.e. in this case \begin{equation} m(\lambda \xi_1, \lambda\xi_2,\lambda\xi_3)=m(\xi_1,\xi_2,\xi_3), \end{equation}by differentiation we get that for any multi index $\alpha \in \mathbb{Z}_{\geq 0}^3$ it holds that \begin{equation} \lambda^{|\alpha|}\partial^\alpha m(\lambda \xi_1, \lambda\xi_2,\lambda\xi_3)=\partial^\alpha m(\xi_1,\xi_2,\xi_3), \end{equation} Then by picking $\lambda = (|\xi_1|^2+|\xi_2|^2+|\xi_3|^2)^{-1/2}$, it follows that \begin{equation} |\partial^\alpha m( \xi_1, \xi_2,\xi_3)| \leq (|\xi_1|^2+|\xi_2|^2+|\xi_3|^2)^{-|\alpha|/2} (\sup_{|\xi|=1}|\partial^\alpha m (\xi)) | \leq c_\alpha |\xi_1|^{-\alpha_1} |\xi_2|^{-\alpha_2} |\xi_3|^{-\alpha_3}. \end{equation} Therefore $m$ satisfies the hypothesis of Corollary 6.2.5 of the same reference, and hence it is a bounded operator on $L^p(\mathbb{R^3}), p>1$.