I want to show that that $u(x,y)=xy\ln|\ln\sqrt{x^2+y^2}|$ is $C^1(\Omega)$ and $H^2(\Omega)$, but not $C^2(\Omega)$, where $\Omega=B_{\frac{1}{2}}(0)$ is the ball centered at $0$ and with radius $1/2$.
To prove that $u$ is $C^1$ I computed explicitly and used polar coordinates for $r\rightarrow 0,$ no problems.
The second derivatives exists, and are given by $$ \frac{\partial^2u}{\partial x^2}=\frac{3xy}{(x^2+y^2)\ln\sqrt{x^2+y^2}}-\frac{x^3y}{(x^2+y^2)^2\ln^2\sqrt{x^2+y^2}}-\frac{2x^3y}{(x^2+y^2)\ln\sqrt{x^2+y^2}}. $$ How can I prove that this function is not continuous at $0$? And how can I prove that this is a weak derivative of for $u$ and belongs to $L^2$? Is there a way that does not envolves using directly the definition?