You can do it the way described in the previous answer, no question, but technically, once you knew the result about $D_1D_2$ and $D_1^2,D_2^2$ and also knew that the support does not matter as long as it is compact, you could just observe that if you change $u(x_1,x_2)$ to $u(ax_1,a^{-1}x_2)$, you'll get
$$
\|D_1D_2u\|_p^p\le C(a^{2p}\|D_1^2u\|_p^p+a^{-2p}\|D_2^2u\|_p^p)\,,
$$

so we actually have
$$
\log \|D_1D_2u\|_p\le C+\frac12[\log\|D_1^2u\|_p+\log\|D_2^2u\|_p]\,.
$$
In general, the same trick and the 2D second order inequality above (the first one) applied to $D_1^{\alpha_1}\dots D_{i-1}^{\alpha_{i-1}}D_i^{\alpha_i-1}D_{i+1}^{\alpha_{i+1}}\dots D_j^{\alpha_j-1}\dots D_n^{\alpha_n}u$ in the $i,j$ 2D plane with subsequent integration over the other coordinates show that if $\alpha_i,\alpha_j>0$, then
$$
\log\|D^\alpha u\|_p\le C+\frac 12[\log\|D^{\alpha'}u\|_p+\log\|D^{\alpha''}u\|_p]
$$
where the multiindex $\alpha'$ is obtained from $\alpha$ by adding $(-1,1)$ to $(\alpha_i,\alpha_j)$ and $\alpha''$ is obtained by adding $(1,-1)$.

But then the maximum over $\alpha$ of any given fixed order of the expression
$$
\log\|D^\alpha u\|_p+\frac C2\sum_i\alpha_i^2
$$
can be attained only at a pure derivative (when all $\alpha_j$ except one are $0$). End of story.

Of course, this doesn't give you a very good constant, but it is a cheap trickery that you can use without looking in the Grafakos book and such as soon as you know the 2D second order result.