It's Gold!

Let $\phi = \dfrac{1 + \sqrt{5}}{2}$ .

For $\dfrac{\phi - 1}{2}:$

$\phi$ is one root of $x^2 - x - 1 = 0 \implies \phi^2 = \phi + 1 \implies$ $\implies \phi^3 = \phi^2 + \phi = 2\phi + 1 \implies \phi^3 - 1 = 2\phi \implies \dfrac{1}{\phi^3 - 1} = \dfrac{1}{2\phi} = \dfrac{(\dfrac{\sqrt{5} - 1}{2})}{2} = \dfrac{\phi - 1}{2} \implies$

$\dfrac{\phi - 1}{2} = \dfrac{1}{\phi^3 - 1} = \dfrac{\dfrac{1}{\phi^3}}{1 - \dfrac{1}{\phi^3}} = \sum_{j = 1}^{\infty} (\dfrac{1}{\phi^3})^{j} = \boxed{(0.001001001001 ...)_{\phi}}$

For $\dfrac{\phi}{2}:$

$\phi^2 = \phi + 1 \implies \implies \phi^3 = \phi^2 + \phi = 2\phi + 1 \implies \phi^3 - 1 = 2\phi \implies$

$1 - \dfrac{1}{\phi^3} = \dfrac{2}{\phi^2} \implies \dfrac{1}{1 - \dfrac{1}{\phi^3}} = \dfrac{\phi^2}{2} \implies \dfrac{\phi^2}{2} = \sum_{n = 0}^{\infty} (\dfrac{1}{\phi^3})^{n} \implies \dfrac{\phi}{2} = \sum_{n = 0}^{\infty} (\dfrac{1}{\phi})^{3n + 1} = \boxed{(0.100100100100 ... })_{\phi}$

$\therefore$ the correct answer is choice (3) .