Can the 2nd Order Partial Derivative of a Function be Evaluated at (0,0)?

[precondition]

This is supposed to be year1 calculus question but I can't answer it.
If f:R_2-->R is 0 if (x,y)=(0,0) and xy(x_2-y_2)/(x_2+y_2) otherwise then evaluate 2nd order partial derivative DxDyf(0,0) and Dy,Dxf(0,0)
The thing is, I get some complicated looking expression for DxDyf(x,y) and I can't simply put x=0 and y=0 in that expression right? because it gives 0/0... What am I misunderstanding here?

 

[precondition]

or is it that DxDyf(0,0) is simply zero because at (0,0) the function is 0?

 

[StatusX]

The derivative is defined for all (x,y) except (0,0), and its derivative at these points is just what you probably found (using the product rule, etc). If the function is differentiable, then its derivative is continuous, and so you can find the derivative at (0,0) by taking the limit. If not, then the derivative at (0,0) is undefined. If you need to be rigorous, you'll need to go back to the definition of the derivative in terms of limits.