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The question states '...then x: is' 

It should state '..then x:y is'

Given

$\dfrac { 1 }{ x } +y=3$ --- Eqn (1)

$x+\dfrac { 1 }{ y } =2$ ---Eqn (2)

Multiplying Eqn (1) by x and Eqn (2) by y, we get:

$x\left( \dfrac { 1 }{ x } +y \right) =3x\quad \Rightarrow 1+yx=3x$ --- Eqn (3)

$y\left( x+\dfrac { 1 }{ y }  \right) =2y\quad \Rightarrow xy+1=2y$ --- Eqn (4)

Subtracting Eqn (4) and Eqn (5), we get:

$1+yx-1-yx=3x-2y$

$\Rightarrow 0=3x-2y$

$\Rightarrow 3x=2y$

$\Rightarrow \dfrac { x }{ y } =\dfrac { 2 }{ 3 } $

$\therefore x:y=2:3$

Hence the answer is B