Let A be open in $\mathbb{R^2}$ and $f:A \rightarrow \mathbb{R}$ such that $f \in C^2$ and $D_{11}f(x,y)+D_{22}f(x,y)=0 $ for any $x,y \in A$. Prove that if $(x_0,y_0)$ is local extremum of f, then all second order derivatives are zero in $(x_0, y_0).$
Consider the hessian matrix of $f$ in $(x_0,y_0),$ $Hf(x_0,y_0)=\begin{bmatrix}D_{11}f(x_0,y_0) & D_{12}f(x_0,y_0)\\D_{21}f(x_0,y_0) & D_{22}f(x_0,y_0)\end{bmatrix}$.
$(x_0,y_0)$ is a local extremum of $f$ so it is either a local maximum or a local minimum and thus $Hf(x_0,y_0)$ is either semidefinite positive (its eigenvalues are all non negative) or semidefinite negative (its eigenvalues are all non positive).
$D_{11}f(x_0,y_0)+D_{22}f(x_0,y_0)=0 \rightarrow D_{11}f(x_0,y_0)=-D_{22}f(x_0,y_0)$ so:
$det\begin{bmatrix}D_{11}f(x_0,y_0) - \lambda & D_{12}f(x_0,y_0)\\D_{21}f(x_0,y_0) & D_{22}f(x_0,y_0) - \lambda\end{bmatrix} = det\begin{bmatrix}D_{11}f(x_0,y_0) - \lambda & D_{12}f(x_0,y_0)\\D_{21}f(x_0,y_0) & -D_{11}f(x_0,y_0) - \lambda\end{bmatrix}$;
$f \in C^2$ so $D_{12}f(x_0,y_0)=D_{21}f(x_0,y_0)$ and then:
$det\begin{bmatrix}D_{11}f(x_0,y_0) - \lambda & D_{12}f(x_0,y_0)\\D_{21}f(x_0,y_0) & -D_{11}f(x_0,y_0) - \lambda\end{bmatrix}=det\begin{bmatrix}D_{11}f(x_0,y_0) - \lambda & D_{12}f(x_0,y_0)\\D_{12}f(x_0,y_0) & -D_{11}f(x_0,y_0) - \lambda\end{bmatrix}$ =
$(D_{11}f(x_0,y_0)- \lambda)(-D_{11}f(x_0,y_0) - \lambda) - 2D_{12}f(x_0,y_0) = 0$;
$(D_{11}f(x_0,y_0)- \lambda)(-D_{11}f(x_0,y_0) - \lambda) - 2D_{12}f(x_0,y_0) = -D_{11}f(x_0,y_0)^2-\lambda D_{11}f(x_0,y_0)+\lambda D_{11}f(x_0,y_0)+ \lambda^2 - 2D_{12}f(x_0,y_0) = -D_{11}f(x_0,y_0)^2+ \lambda^2 - 2D_{12}f(x_0,y_0) = 0$;
So $\lambda^2 = 2D_{12}f(x_0,y_0)+D_{11}f(x_0,y_0)^2$;
I don't know how to continue from here or if this is the right path. Any help?