Yep, you're right. Basically he means X0 = a. You are taking the series about X0 = 3. I.e:

f(x) = f(3)/0! + f'(3)*(x-3)/1! + f''(3)*(x-3)^2/2! + f'''(3)*(x-3)^3/3! + ...

so in your particular case you have:

f(x) = 1/x, f'(x) = -1/x^2, f''(x) = 2/x^3, f'''(x) = -6/x^4, ... fn(x) = (n!*(-1)^n)/x^(n+1)

which gives

f(x) (about x = a) = 1/a - (1/a^2)*(x-a) + (1/a^3)*(x-a)^2 - (1/a^4)*(x-a)^3 + - + - ...

about x=3 this is:

f(x) = 1/3 - (1/9)*(x-3) + (1/27)*(x-3)^2 - (1/81)*(x-3)^3 + ...

plug y = 1/x and y = 1/3 - (1/9)*(x-3) + (1/27)*(x-3)^2 - (1/81)*(x-3)^3 into your calculator and open the table. Put in values near 3, like 3.001 or 2.999 and you'll find that both y's return nearly the same value.