We have $$\int_{\mathbb R^2} \dfrac{\partial^2 f}{\partial x^2}\dfrac{\partial^2 f}{\partial y^2}dx dy = \int_{\mathbb R}\left(\int_{\mathbb{R}}\dfrac{\partial^2 f}{\partial x^2}\dfrac{\partial^2 f}{\partial y^2}dy\right)dx$$
and
$$\int_{\mathbb{R}}\dfrac{\partial^2 f}{\partial x^2}\dfrac{\partial^2 f}{\partial y^2}dy = \left[\dfrac{\partial^2 f}{\partial x^2}\dfrac{\partial f}{\partial y}\right]_{y=-\infty}^{y=+\infty}-\int_{\mathbb R}\dfrac{\partial^3 f}{\partial x^2\partial y}\dfrac{\partial f}{\partial y}dy $$
hence
$$\int_{\mathbb R^2} \dfrac{\partial^2 f}{\partial x^2}\dfrac{\partial^2 f}{\partial y^2}dx dy =-\int_{\mathbb R}\left(\int_{\mathbb R}\dfrac{\partial^3 f}{\partial x^2\partial y}\dfrac{\partial f}{\partial y}dx\right)dy.$$
To conclude, we notice that
$$\int_{\mathbb R}\dfrac{\partial^3 f}{\partial x^2\partial y}\dfrac{\partial f}{\partial y}dx=\left[\dfrac{\partial^2 f }{\partial x\partial y}\dfrac{\partial f}{\partial y}\right]_{x=-\infty}^{x=+\infty} -\int_{\mathbb R}\dfrac{\partial^2 f }{\partial x\partial y}\dfrac{\partial^2 f }{\partial y\partial x}.$$
Added later: of course, the brackets vanish since $f$ and its partial derivatives have a compact support.