what's the ordinary derivative of the kronecker delta function?

Let's approach this problem using the Heaviside step function, $$H(x) = \begin{cases} 0, & x < 0 \\ 1, & x \ge 0. \end{cases}$$ Notice we choose the convention that $H(0) = 1$.

The Kronecker delta is then $$\delta_{m n} = H(m-n) + H(n-m) - 1.$$ The derivative of the Heaviside step function is the Dirac delta function, $H'(x) = \delta(x)$. Therefore, $$\begin{eqnarray*} \frac{\partial}{\partial m} \delta_{m n} &=& \delta(m-n) - \delta(n-m) \\ &=& 0 \end{eqnarray*}$$ since $\delta(-x) = \delta(x)$.

Addendum: If by ordinary derivative it is meant the derivative in the classical sense then of course the derivative is everywhere zero, except at $m = n$ where it is undefined. In order to define it at $m=n$, which seems to be the spirit of the question, we must go to the distributional derivative in which case we find the derivative is $0$.

I take it you are defining $\delta(x,y)$ to be a function of 2 real variables given by $1$ if $x=y$ and $0$ otherwise. I suggest then that you go to the definition of the derivative for functions of two variables and see what happens when you apply it to this function, and then report back to us on your findings.

As Arturo wrote in a comment, the function $f:\mathbb R\to\mathbb R$ such that for all $x\in\mathbb R$ we have $$f(x)=\begin{cases}0, & \text{if $x\neq0$;} \\1, & \text{if $x=0$;}\end{cases}$$ is not continuous at $0$. It follows that it does not have a derivative at $0$. On the other hand, a trivial calculation using the definition itself of what a derivative is will show that it has a derivative at all other points, which is zero.

N.B. Of course, the above paragraph is only true if we interpret your question in classical terms. There are several other interpretations we could give to your question, and that would change the answer. We could think that you really are talking about derivatives in the sense of theory of distributions, as in oenamen's answer. You could have something else in mind. Please be more explicit.