For a real number $y$, consider $[y]$ denotes the greatest integer less than or equal to $y$. If $f(x) = \dfrac{\tan(\pi[x - \pi])}{1 + [x]^2}$, then

Given the function $f(x) = \dfrac{\tan(\pi[x - \pi])}{1 + [x]^2}$.

We know that for any real number $x$, the value of $[x - \pi]$ is always an integer. Let $[x - \pi] = k$, where $k \in \mathbb{Z}$.

The numerator of the function becomes $\tan(k\pi)$. Since the tangent of any integer multiple of $\pi$ is exactly zero, we have $\tan(k\pi) = 0$ for all $x \in \mathbb{R}$.

For the denominator, since $[x]$ is an integer, $[x]^2 \ge 0$, which implies $1 + [x]^2 \ge 1$. Thus, the denominator is never zero and is well-defined for all real $x$.

Therefore, the function simplifies to:

$f(x) = \dfrac{0}{1 + [x]^2} = 0$ for all $x \in \mathbb{R}$.

Since $f(x)$ is a constant function, its derivative is zero everywhere:

$f'(x) = 0$ for all $x \in \mathbb{R}$.

Hence, $f'(x)$ exists for all $x$.

Answer: $f'(x)$ exists for all $x$