The value of $\lim\limits_{x \to 3}\left[\dfrac{1}{x-3} + \dfrac{9x}{27-x^3}\right]$ is:

The given limit is $\lim\limits_{x \to 3}\left[\dfrac{1}{x-3} + \dfrac{9x}{27-x^3}\right]$

Using the identity $a^3 - b^3 = (a-b)(a^2+ab+b^2)$, we can write $27 - x^3 = -(x^3 - 27) = -(x-3)(x^2+3x+9)$

Substituting this into the limit expression:

$\lim\limits_{x \to 3}\left[\dfrac{1}{x-3} - \dfrac{9x}{(x-3)(x^2+3x+9)}\right]$

Taking the common denominator:

$= \lim\limits_{x \to 3}\left[\dfrac{x^2+3x+9-9x}{(x-3)(x^2+3x+9)}\right]$

$= \lim\limits_{x \to 3}\left[\dfrac{x^2-6x+9}{(x-3)(x^2+3x+9)}\right]$

$= \lim\limits_{x \to 3}\left[\dfrac{(x-3)^2}{(x-3)(x^2+3x+9)}\right]$

Since $x \to 3$, $x \neq 3$, we can cancel the common factor $(x-3)$ from the numerator and the denominator:

$= \lim\limits_{x \to 3}\left[\dfrac{x-3}{x^2+3x+9}\right]$

Substituting $x = 3$ into the simplified expression:

$= \dfrac{3-3}{3^2+3(3)+9}$

$= \dfrac{0}{27} = 0$

Answer: $0$