I tried to practice haskell a bit but I didn't get the following code to work:
rems :: Int -> [Int] -> [Int]
rems _ [] = []
rems k (x:xs)
| k == x = rems k xs
| otherwise [x] ++ rems k xs
main = print $
rems 3 [5, 3, 2]
This function removes every k from a list x if you call it rems k x. I know that it should be working because we wrote it down in university.
I'm using this IDE: https://repl.it/languages/haskell
It might be useful to know that otherwise isn't a keyword or special symbol like =, it's actually simply a boolean value defined in the prelude as True. I.e., a guard with otherwise works syntactically the same as any other guard, just the condition is trivial. You might also write it
rems _ [] = []
rems k (x:xs)
| k == x = rems k xs
| True = [x] ++ rems k xs
The alignment I chose above (all = aligned and at least two† spaces away from the guard-condition) is completely a matter of taste, but I think it helps avoiding confusion like the one you've found yourself in.
BTW, [x] ++ ... can be shortened to x : .... The preferred form of writing the function is
rems _ [] = []
rems k (x:xs)
| k==x = rems k xs
| otherwise = x : rems k xs
†Most people align the =s but use only one space. That's ok, but IMO k == x = res looks deceptive with all those equals-characters, and k and x further away from each other than from the result. k==x = res seems tidier.