I'm afraid neither is correct.
To add all fractions together, you need to make the denominators equal.
For the first term with A this means:
A / (x-1) = A(x-2)(x-3) / [(x-1)(x-2)(x-3)].
if a equation is 3x/(x-1)(x-2)(x-3) = A/(x-1) + B/(x-2) + C/(x-3)
will the next step is
= A(x-3) + B(x-1)(x-2) + C(x-1)
= A(x-3) + B(x-1)(x-3) + C(x-1)
You have the nominators correct now, but you should leave out the denominators.
You have multiplied the left hand side with (x-1)(x-2)(x-3) to give you 3x.
The same should be done on the right hand side, meaning there would be no denominators left.