[(3/2)^2-1/2] : (1/3*3/2+3)=
(9/4-1/2):(1/2+3)=
(9-2)/4 : (1+6)/2=
7/4:7/2=
7/4*2/7 = 1/2
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$\dfrac{x^2-y}{\dfrac{1}{3}x+3}=$
$=\dfrac{\left(\dfrac{3}{2}\right)^2-\dfrac{1}{2}}{\dfrac{1}{\cancel3_1}·\dfrac{\cancel3^1}{2}+3}=$
$=\dfrac{\dfrac{9}{4}-\dfrac{1}{2}}{\dfrac{1}{1}·\dfrac{1}{2}+3}=$
$=\dfrac{\dfrac{9-2}{4}}{\dfrac{1}{2}+3}=$
$=\dfrac{\dfrac{7}{4}}{\dfrac{1+6}{2}}=$
$=\dfrac{\dfrac{7}{4}}{\dfrac{7}{2}}=$
$=\dfrac{7}{4}·\dfrac{2}{7}=$
$=\dfrac{\cancel7^1}{\cancel4_2}·\dfrac{\cancel2^1}{\cancel7_1}=$
$= \dfrac{1}{2}·\dfrac{1}{1}=$
$= \dfrac{1}{2}$