The best thing to do is work one column at a time, working from left to right. You want to get a "1" at the pivot position (the main diagonal element for that particular column) so divide the entire row by that number. (Unless, of course, it is 0. If so, swap with a lower row. If the pivot and all lower numbers in that column are 0, the matrix is not invertible). Then, to get a 0 in that column in rows both above and below that pivot row, subtract that number times the pivot row from the row.
Here, you have
Since there is already a 1 in the pivot position (first row first column) you don't have to do anything to the first row (divide by 1). There is a 1 in second row, first column so subtract (1 times) the first row from the second column. There is already a 0 in the first column of the third row so you don't have to do anything to the third row (subtract 0 times the first row from the third row).
Now, we have a 0 in the pivot position (second row of the second column) so we have to swap the second and third rows
Since there is a 2 in the pivot position now, divide the second row by 2. Since there is a 1 in the first row of the second column, subtract 1 times the new second row from the first row. Since there is a 0 in the third row of the second column, subtract 0 times the new second row from the first row (in other words, don't change it).
The pivot position is now the third row of the third column. There is a 1 there so we don't have to do anything to the third row. There is a in the first row of the third column so we add the third row to the first row. There is a in the second row of the third column so we subtract the third row from the second row.
That is, the inverse to is