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...