differentiation of fractional part of $x$

Your function $\{x\}$ has derivative $1$ as you note, except $\{x\}$ has jumps (of $-1$) at each integer. In the theory of distributions, the derivative of a unit jump at $0$ is a measure called $\delta$. So $$ \frac{d}{dx}\{x\} = 1 - \sum_{n \in \mathbb Z} \delta(x-n) \tag{1}$$ (This is a simple example of a Lebesgue decomposition of a signed measure.) What does it mean? For example, we may write a Stieltjes integral like this $$ \int_{-\infty}^{+\infty} \varphi(x)\;d\{x\} = \int_{-\infty}^{+\infty} \varphi(x)\;dx - \sum_{n \in \mathbb Z} \varphi(n) \tag{2}$$ for nice enough functions $\varphi$. For example, this works when $\varphi$ is continuous with compact support.

Is the function continuous? No

So the derivative is undefined for $x \in \mathbb{Z}$ and $1$ everywhere else.

It is not possible to differentiate the fractional part of x when $x \in \mathbb{Z}$. This is because the graph of $\{x\}$ is not continuous. So its derivative does not exist.

Why its derivative does not exist? If you look at the right-hand derivative and left-hand derivative of the integral values of $x$, they are not the same. For a function to be differentiated, the left-hand derivative and right-hand derivative must be the same. Hence, its derivative does not exist at $x \in \mathbb{Z}$.

However, other than the integral values of $\{x\}$ the derivatives exist. The derivative for other values of $x$ is evidently 1. To generalise the derivative of ($\frac{d}{dx}\{x\}, x \in \mathbb{R-Z} = 1$).