As promised, here is my walkthrough for this puzzle
While making the walkthrough, I found out that it's a pretty nasty one! Sorry for that!
1.
Realise the sum in every row, column and nonet is 28
2.
R12C3 = {1,2}
3.
R89C3 = {0,4}
4.
R456C3 = {5,6,7}
5.
R7C789 sum = 15 + R7C789 sum = 11 means R168C8 sum = 2 ={0,0,2}
R1C8 = 2
R68C8 = 0
6.
R7C89 = {1,6}
7.
R23C8 = {5,7}
8.
1 in nonet must be R5C2 or R5C4
R5C3 = 5
9.
(4)8 in R5
R5C124 = {0,1,2}
10.
R9C8 = 3
11.
R9C79 = {5,7}
12.
R1C3 = 1
R2C3 = 2
13.
2 in nonet must be in R4C1 or R4C9
14.
2 in nonet must be in RR9C2
15.
(3)10 in C4 must contain a 7 without 2
R123C4 = {0,3,7}
16.
(3)6 in C4 = {0,1,5}
R7C4 = 0
R8C4 = 5
R9C4 = 1
17.
R5C2 = 1
R5C4 = 2
R5C1 = 0
18.
R8C3 = 4
R9C3 = 0
R7C2 = 0
19.
R46C4 = {4,6}
20.
(4)16 in C2 + 1 (R5C2) + 0 (R7C2) + 2 (R9C2) = 19
sum R68C2 = 9 = {3,6}
21.
R3C1 = 6
R1C1 = 3
22.
R3C8 = 7
R2C8 = 5
23.
5 in nonet must be in R46C7
R9C7 = 7
R2C8 = 5
24.
6 in rows 4&6 in columns 3&4
R6C2 = 3
R8C2 = 6
25.
R4C8 cannot be a 6
6 in nonet in R125C9
R7C9 = 1
R7C8 = 6
26.
R5C8 = 4
R4C8 = 1
27.
R3C4 + R3C5 have a sum of 4 inbetween (3)11 and (4)13
R3C4 = 3
R3C5 =1
28.
R3C6 = 0
R3C9 = 4
29.
R29C5 = {4,6}
Because of (4)9 in C5: R9C5 = 4 and R2C5 = 6
30.
R9C1 = 0
R9C6 = 6
R1C9 = 6
R5C7 = 6
R6C1 = 4
R6C4 = 6
R4C4 = 4
R6C3 = 7
R4C3 = 6
R6C7 = 1
R4C7 = 1
R4C7 = 5
R6C7 = 1
R6C9 = 2
R4C1 = 2
R7C5 = 2
31.
(4)9 in C5 = {0,2,3,4}
R6C5 = 0
R8C5 = 3
And the rest are all easy singles...
