I am told to find $f_{xy}$ in the distributional sense, where: $$f(x,y) = \begin{cases} 1 & y\geq x^3\\ 0 & y<x^3 \end{cases}$$ Now I know that the pointwise derivatives $f_{xy}$ and $f_{yx}$ are zero everywhere. Calculating the second derivative in the distributional sense where $\phi\in C^{\infty}_c (\mathbb R^2)$:
$$\langle f_{xy},\phi\rangle = \langle f,\phi_{xy}\rangle = \iint_{\mathbb R^2}f(x,y)\cdot \phi_{xy} dx dy = \int_{-\infty}^\infty \int_{x^3}^\infty \phi_{xy}(x,y) dy dx$$ Integrating: $$-\int_{-\infty}^\infty \phi_x(x,x^3) dx = \phi(x,x^3)\bigg|_{-\infty}^\infty =0$$ As $\phi$ is of compact support. Is my reasoning correct? Am I missing anything?
The solution is correct up to $$-\int_{-\infty}^\infty \phi_x(x,x^3) dx = \phi(x,x^3)\bigg|_{-\infty}^\infty$$ which is false because $$ \phi_x(x,x^3) \ne \frac{d}{dx}(\phi(x,x^3)) $$ On the left, we first take the derivative and then plug $x^3=y$. On the right, we plug $y=x^3$ and then take the derivative.
The expression $-\int_{-\infty}^\infty \phi_x(x,x^3) dx $ does not simplify further. Since the evaluation of $-\phi_x$ is the $x$-derivative of Dirac delta, one can express the distribution $u_{xy}$ as $$ u_{x,y}= \int_{-\infty}^\infty \frac{\partial \delta}{\partial x} (x-t, y-t^3)\,dt $$ but I'd probably leave it at $$ u_{x,y} = \left(\phi\mapsto -\int_{-\infty}^\infty \phi_x(x,x^3) dx\right) $$ which is more explicit.