Proof that the derivative of a constant is zero

The "$x^0$" really plays no role; besides, it is $1$ and any other constant works with the same proof.

In any case, as you write, if $f(x)=c$ for some number $c$, then $$ \frac{f(x+h)-f(x)}h=\frac{c-c}h=0. $$ So, the Newton quotient is already zero before taking the limit as $h\to0$.

Keep in mind that "limit" has a precise, formal meaning: $$\lim_{x\rightarrow c}f(x)=L \quad\iff\quad \forall \epsilon>0\exists \delta>0\forall x[0<\vert x-c\vert<\delta\implies \vert f(x)-L\vert<\epsilon].$$ In words, the limit as $x$ approaches $c$ of $f(x)$ is $L$ if, as we approach (but not reach - this is the "$0<$" clause) $c$, the value of $f$ approaches $L$.

For any $h$ other than $0$, we have ${0\over h}=0$; so indeed $\lim {0\over h}=0$. The fact that, at $h=0$, the expression $0\over h$ is undefined - and as a consequence, doesn't equal $0$ - doesn't effect the limit; it just means that the function isn't as nicely behaved at $h=0$ as it could be.