$\text { The number of points of discontinuity of the rational function } f(x)=\dfrac{x^2-3 x+2}{4 x-x^3}$

The function is defined as $f(x) = \dfrac{x^2 - 3x + 2}{4x - x^3}$.

Factorizing the numerator and the denominator:

Numerator: $x^2 - 3x + 2 = (x - 1)(x - 2)$

Denominator: $4x - x^3 = x(4 - x^2) = x(2 - x)(2 + x) = -x(x - 2)(x + 2)$

Thus, $f(x) = \dfrac{(x - 1)(x - 2)}{-x(x - 2)(x + 2)}$.

The function is undefined where the denominator is zero, which occurs at $x = 0$, $x = 2$, and $x = -2$.

Simplifying the expression for $x \neq 2$:

$f(x) = \dfrac{x - 1}{-x(x + 2)}$ for $x \neq 2$.

At $x = 0$ and $x = -2$, the denominator is zero while the numerator is non-zero, so these are points of infinite discontinuity.

At $x = 2$, the limit exists:

$\lim_{x \to 2} f(x) = \lim_{x \to 2} \dfrac{x - 1}{-x(x + 2)} = \dfrac{2 - 1}{-2(2 + 2)} = \dfrac{1}{-8} = -\dfrac{1}{8}$.

Since the limit exists but the function is not defined at $x = 2$, $x = 2$ is a point of removable discontinuity.

Therefore, the function is discontinuous at $x = 0$, $x = 2$, and $x = -2$. The total number of points of discontinuity is 3.

Answer: 3