Solution of equations $\dfrac{2}{x} - \dfrac{3}{y} = 13 \text{ and } \dfrac{3}{x} + \dfrac{2}{y}$ = 0 is :

1. $x = \dfrac{1}{2}, y = \dfrac{1}{3}$

2. $x = -\dfrac{1}{2}, y = \dfrac{1}{3}$

3. $x = \dfrac{1}{2}, y = -\dfrac{1}{3}$

4. $x = -\dfrac{1}{2}, y = -\dfrac{1}{3}$

Given, equations :

$\dfrac{2}{x} - \dfrac{3}{y} = 13$ …….(1)

$\dfrac{3}{x} + \dfrac{2}{y} = 0$ ……..(2)

Multiplying equation (1) by 2, we get :

$\Rightarrow 2\Big(\dfrac{2}{x} - \dfrac{3}{y}\Big) = 2 \times 13 \\[1em] \Rightarrow \dfrac{4}{x} - \dfrac{6}{y} = 26 \text{ …….(3)}$

Multiplying equation (2) by 3, we get :

$\Rightarrow 3\Big(\dfrac{3}{x} + \dfrac{2}{y}\Big) = 3 \times 0 \\[1em] \Rightarrow \dfrac{9}{x} + \dfrac{6}{y} = 0 \text{ …….(4)}$

Adding equations (3) and (4), we get :

$\Rightarrow \Big(\dfrac{4}{x} - \dfrac{6}{y}\Big) + \Big(\dfrac{9}{x} + \dfrac{6}{y}\Big) = 26 + 0 \\[1em] \Rightarrow \dfrac{4}{x} + \dfrac{9}{x} - \dfrac{6}{y} + \dfrac{6}{y} = 26 \\[1em] \Rightarrow \dfrac{13}{x} = 26 \\[1em] \Rightarrow x = \dfrac{13}{26} = \dfrac{1}{2}.$

Substituting value of x in equation (2), we get :

$\Rightarrow \dfrac{3}{\dfrac{1}{2}} + \dfrac{2}{y} = 0 \\[1em] \Rightarrow 6 + \dfrac{2}{y} = 0 \\[1em] \Rightarrow \dfrac{2}{y} = 0 - 6 \\[1em] \Rightarrow \dfrac{2}{y} = -6 \\[1em] \Rightarrow y = \dfrac{2}{-6} = -\dfrac{1}{3}.$

Hence, Option 3 is the correct option.