Let $f\in S(\mathbb{R}^2)$ be a function in Schwartz space. This means that $f\in C^{\infty}(\mathbb{R}^2)$ and for all multi-indices $\alpha$ and integers $N$ there exists $C_{N,\alpha}$ such that, $$|\partial_{x}^{\alpha}f(x)|\leq C_{N,\alpha}(1+|x|)^{-N}.$$
We want to show that $$||f_{xy}||_{L^2}\leq 2 ||\Delta f||_{L^{2}}$$ where $f_{xy}$ is the partial derivative of $f$ with respect to $x$ first and then $y.$
Here is what I tried. First, we must observe that $S(\mathbb{R}^2)\subset L^2(\mathbb{R}^2)$ otherwise taking norms in $L^2$ does not make sense. Next, we compute, $$\int \int f_{xy}f_{xy}dxdy\leq 4\left(\int\int f_{xx}^2 dxdy+\int \int f_{yy}^2dxdy + \int \int 2f_{xx}f_{yy}dxdy\right).$$ I think there is integration by parts required here, but I am not sure how to proceed after this. Any hints will be much appreciated.
$\newcommand\norm[1]{\left\lVert#1\right\rVert}$
First observe that, $$\norm{f_{xy}}_2 = \norm{xy\hat{f}}_2.$$ To see this, let $g = f_{xy}$, then \begin{align*} \hat{g}(\zeta_1, \zeta_2) &= \int \int e^{-i\zeta_1x-i\zeta_2y}f_{xy}(x,y)dxdy\\ &= (-i\zeta_1)\int \int e^{-i\zeta_1x-i\zeta_2y}f_{y}(x,y)dxdy\\ &= \zeta_1 \zeta_2 \int \int e^{-i\zeta_1x-i\zeta_2y}f(x,y)dydx\\ &= \zeta_1\zeta_2 \hat{f}. \end{align*} So $\norm{f_{xy}}_2 =\norm{xy\hat{f}}_{2}.$ Now we consider \begin{align*} h &=\Delta f = f_{xx} + f_{yy} \implies \\ \hat{h}(x,y) &= (x^2 + y^2) \hat{f}(x,y). \end{align*} Thus finally, we use the inequality $2|xy|\leq x^2+y^2$ to get that, $$\norm{f_{xy}}_2 =\norm{xy\hat{f}}_2\leq 2\norm{(x^2+y^2)\hat{f}}_2\leq 2\norm{\Delta f}_2.$$