#f(x) = x/(x^2 - 3 x + 2) = x/((x-1)(x-2)) = A/(x-1)+B/(x-2)#

#(A(x-2)+B(x-1))/((x-1)(x-2))=((A+B)x-(2A+B))/((x-1)(x-2))#

Solving for #A,B#

#{
(A+B=1),
(2A+B=0)
:}#

we obtain #A=-1,B=2# So

#f(x) = -1/(x-1)+2/(x-2)#

Now considering that #(x^{n+1}-1)/(x-1)=1+x+x^2+cdots+x^n#

if #abs(x) < 1# we know

#lim_{n->oo}(x^{n+1}-1)/(x-1) = -1/(x-1) = sum_{n=0}^oo x^n#

Using those results and supposing that #abs(x) < 1# we propose

#S(x) = -1/(x-1)+2/(x-2) = sum_{n=0}^oo x^n-sum_{n=0}^oo (x/2)^n#

so #S(x) = sum_{n=0}^oo(1-1/2^n)x^n#.

This serie is convergent to #f(x)# for #abs(x) < 1#

Supposing you need a series representation for #f(x)# in the surroundings of #x=3# you can proceed as follows.

Making #y = x-3# then

#f(y) = -1/(y+2)+2/(y+1)#

so with #abs(y) < 1#

#f(y)=lim_{n->oo}{-(1-(-y/2)^n)/(2(1-(-y/2)))+2(1-(-y)^n)/(1-(-y))}#

or

#f(y) = -1/2 sum_{n=0}^oo(-y/2)^n+2sum_{n=0}^n(-y)^n#

Finally

#S(x) = sum_{n=0}^oo (-1)^n(2-1/2^{n+1})(x-3)^n#

This series converges to #f(x)# for #2 < x < 4#