The sum of $-\dfrac{1}{3}$ and its multiplicative inverse is

1. 0

2. -3

3. $-1\dfrac{2}{3}$

4. $-3\dfrac{1}{3}$

Multiplicative inverse of $-\dfrac{1}{3}$ is -3.

Sum = $-\dfrac{1}{3} + (-3) = -\dfrac{1}{3} - \dfrac{3}{1}$

L.C.M. of 3 and 1 is 3.

Now, expressing each fraction with denominator 3:

$= -\dfrac{1 \times 1}{3 \times 1} - \dfrac{3 \times 3}{1 \times 3} \\[1em] = -\dfrac{1}{3} - \dfrac{9}{3} \\[1em] = \dfrac{-1 - 9}{3} \\[1em] = \dfrac{-10}{3} \\[1em] = -3\dfrac{1}{3}$

Hence, option 4 is the correct option.