Let $f ( x )=2 x +3 y$
$f ( x )=2 x +\frac{18}{ x } $
$(\because xy =6$ given $)$
On differentiating, we get
$f' x=2-\frac{18}{x^{2}}$
Put $f '( x )=0$ for maximum or minima.
$\Rightarrow 0=2-\frac{18}{x^{2}} $
$\Rightarrow x=\pm 3$
And $f '' x =\frac{36}{ x ^{3}}$
$\Rightarrow f '' 3=\frac{36}{3^{3}}>0$
$\therefore $ At $x =3$, If $x$ is minimum.
The minimum value is
$f(3)=2(3)+3(2)=12$