Suppose ddz(f(z)) exists at a point z. Is ddz(f(z)) continuous at z?

Intuitively, you would maybe think so but the answer is no.
As a counterexample, consider
$g(x):=\{\begin{array}{ll}{x}^2\sin (1/x),& x\ne 0\\ 0& x=0\end{array}$
Then the derivative is
$g^{\prime }(x)=\{\begin{array}{ll}2x^2\sin (1/x)-\cos (1/x),& x\ne 0\\ 0& x=0\end{array}$
Note that ${\displaystyle \underset{x\to 0}{lim}{g}^{\prime }\left(x\right)}$ does not exist, even though ${\displaystyle g^{\prime }\left(0\right)=0}$.

If you are thinking to functions over the complex numbers, it is a celebrated result of Goursat that if the derivative exists over an open set, then the function is analytic over the open set; in particular its derivative is analytic as well, in particular differentiable and, therefore, continuous.
However, the derivative might not exist over an open set. An example is the function ${\displaystyle f\left(z\right)={\left|z\right|}^2=z\bar{z}}$, which is only differentiable at zero; indeed, the Wirtinger derivative is
${\displaystyle \frac{\partial f}{\partial \bar{z}}=z}$
and a function can be differentiable (over the complex numbers) only where the Wirtinger derivative vanishes (it’s just the Cauchy-RIemann condition expressed in a different way). The derivative actually exists at zero, because
${\displaystyle \underset{z\to 0}{lim}\frac{f\left(z\right)-f\left(0\right)}{z-0}=\underset{z\to 0}{lim}\bar{z}=0}$
The derivative only exists at a point so it is continuous over its domain, but I guess it’s not the notion of continuity you were thinking to.