Arrange numbers 1 to 9 into the upsilon grid

Arrange numbers 1 to 9 into the octagon, so the operation is correct.
C is a constant.

Do the math operation in sequence, ($×$) and ($/$) is NOT HIGHER than ($+$) and ($-$).

Answer

You could deduce things using the equations.

As mentioned by @Bojan B (with different notation), we can write down some equations:
(a - b) * c = C
(d * e) / f = C
(g - h) + i = C
(a + d) / g = C

(b + e) / h = C
(c * f) - i = C

Then we can start writing down knowns:
1. a != b != c != d != e != f != g != h != i
2. C is positive, because (d * e) / f will not produce less than 1.
3. C is an integer because (g - h) + i will not produce a fraction.
4. a > b. Otherwise, (a - b) * c produces a negative value and contradicts #2.
5. a != 1. Otherwise, b is not in our set of values. b != 9. Otherwise, a is not in our set of values.
6. If a + d is a prime number, then g = 1.
7. If b + e is a prime number, then h = 1.
8. a + d and b + e both cannot be prime, because both g and h would be 1. Contradicts #1.

9. g > h and/or i > g - h. Otherwise, (g - h) + i produces a negative and contradicts #2.
10. f != 5 and f != 7 because d or e would need to be 5 or 7 to satisfy (d * e) / f and contradict #1.
11. C < 17 because the largest value of (g - h) + i is (9 - 1) + 8 = 16.
12. c * f < 25 because (c * f) - i = C or (c * f) = C + i and the largest possible value of C + i is 16 + 9 = 25 (#11). In order for cf = 25, c = f = 5 (contradicts #1).
13. c * f multiplies to 2, 3, 4, 6, 8, 9, 12, 16, 18, or 24 (no multiplies of 5 or 7 because #10).
14. c * f > i. Otherwise, (c * f) - i produces 0 or less.

One could keep going with this strategy and (hopefully) eventually deduce a non-brute force solution.

Edit: continuing my efforts.

Is it possible to add two equations together if we aren't following BODMAS?

(g - h) + i = C
+ (c * f) - i = C
= (g - h) + (c * f) = 2C
which would tell us both g - h and c * f are multiples of 2. So g and h are both even or both odd, and from #13, c * f no longer multiplies to 3 or 9.
Also c * f > g - h because c * f = (g - h) + 2i.

We also have 3 ratios that equal C: (d * e) / f, (a + d) / g, and (b + e) / h. Ratios that could potentially satisfy these are:
1 = 6/6, 8/8, 9/9
2 = 2/1, 4/2, 6/3, 8/4, 12/6, 16/8, 18/9
3 = 3/1, 6/2, 9/3, 12/4, 18/6, 24/8
4 = 4/1, 8/2, 12/3, 16/4, 24/6
6 = 6/1, 12/2, 18/3, 24/4

8 = 8/1, 16/2, 24/3
I've excluded ones where the numerator can only be factored by the denominator. Since we need 3 unique denominators in a set, I also excluded sets with less than 3 ratios.
So we at least know C is 1, 2, 3, 4, 6 or 8.