Finding a Second Partial Derivative in the Distributional Sense 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.
