The second order euler constant is k times k minus 1 not k squared

1. A.$3\partial f/\partial x_1(1,1)+3\partial f/\partial x_2(1,1)=f(1,1)$
2. B.$\partial f/\partial x_1(1,-1)-\partial f/\partial x_2(1,-1)=3f(1,-1)$✓
3. C.$x_1^2{\partial }^{2}f/\partial x_1^2(x_1,x_2)+x_2^2{\partial }^{2}f/\partial x_2^2(x_1,x_2)+2x_1{x}_2{\partial }^{2}f/\partial x_1\partial x_2(x_1,x_2)=6f(x_1,x_2)$✓
4. D.$x_1^2{\partial }^{2}f/\partial x_1^2(x_1,x_2)+x_2^2{\partial }^{2}f/\partial x_2^2(x_1,x_2)+2x_1{x}_2{\partial }^{2}f/\partial x_1\partial x_2(x_1,x_2)=9f(x_1,x_2)$

The trap it tests

Execution slip

The idea was right. The computation was not.