Brain works

Using Sine Rule we know that:

$\dfrac {a}{\sin{A}} = \dfrac {b}{\sin{B}} \quad \Rightarrow \dfrac {\sin{A}}{\sin{b}} = \dfrac {a}{b} = \dfrac {6}{3} = 2\quad \Rightarrow \sin{A} = 2\sin{B}$

Let $\quad \sin{B} = x \quad \Rightarrow \cos{B} = \sqrt{1-x^2}$

$\quad \Rightarrow \sin{A} = 2x \quad \Rightarrow \cos{A} = \sqrt{1-4x^2}$

$\cos{A-B} = \cos{A}\cos{B}+\sin{A}\sin{B} = \dfrac {4}{5}$

$\Rightarrow \sqrt{1-4x^2}\sqrt{1-x^2}+(2x)(x) = \dfrac {4}{5}$

$\Rightarrow \sqrt{1-4x^2}\sqrt{1-x^2} = \dfrac {4}{5}-2x^2$

$\Rightarrow (1-4x^2)(1-x^2) =\left( \dfrac {4}{5}-2x^2\right)^2$

$\Rightarrow 1-5x^2 + 4x^2 = \dfrac {16}{25}-\dfrac {16}{5}x^2+4x^4$

$\Rightarrow \left(5 -\dfrac {16}{5} \right) x^2 = 1 - \dfrac {16}{25} \quad \Rightarrow \dfrac {9}{5}x^2 = \dfrac {9}{25} \quad \Rightarrow x = \dfrac {1}{\sqrt{5}}$

$\Rightarrow \sin{B} = \frac {1}{\sqrt{5}} \quad \Rightarrow \cos{B} = \frac {2}{\sqrt{5}} \quad \Rightarrow \sin{A} = \frac {2}{\sqrt{5}} \quad \Rightarrow \cos{A} = \dfrac {1}{\sqrt{5}}$

The area of $\triangle ABC$ is given by:

$A = \frac{1}{2}(3\cos{A}+6\cos{B})(6\sin{B}) = \dfrac{1}{2}\left( \dfrac {3}{\sqrt{5}} + \dfrac {12}{\sqrt{5}} \right) \left( \dfrac {6}{\sqrt{5}} \right)$

$\quad = \dfrac{1}{2}\times \dfrac {13}{\sqrt{5}} \times \dfrac {6}{\sqrt{5}} = \dfrac {45}{5} = \boxed{9}$