const arrayOne = [
  { id: 1, name: "one" },
  { id: 2, name: "two" },
  { id: 3, name: "three" },
];
const arrayTwo = [
  { id: 2, name: "two" },
  { id: 3, name: "three" },
];

const arrayTwoIds = new Set(arrayTwo.map((el) => el.id));
const arrayOneFiltered = arrayOne.filter((el) => !arrayTwoIds.has(el.id));

console.log(arrayOneFiltered);
// [ { id: 1, name: 'one' } ]

Depending on the size of the array, creating a set can improve performance, as you do not need to loop over arrayTwo arrayOne.length times but only once. After that, you can look up the existence of an id in arrayTwo in constant time.

Yet, as pointed out in another answer, this is not necessary if the arrays are small (like in your example). In this case, you could also use this one-liner:

arrayOne = arrayOne.filter((elOne) => !arrayTwo.some((elTwo) => elOne.id === elTwo.id));

Here, arrayOne would need to be mutable, i.e. defined with let.

You can find it by comparing it with id.

And arrayOne must be a let.

   let arrayOne = [{id: 1, name: 'one'}, {id: 2, name: 'two'}, {id: 3, name: 'three'}];
   const arrayTwo = [{id: 2, name: 'two'}, {id: 3, name: 'three'}];
   
   arrayOne = arrayOne.filter(one => !arrayTwo.find(two => one.id == two.id));
   
   console.log(arrayOne);

const arrayOne = [{
  id: 1,
  name: 'one'
}, {
  id: 2,
  name: 'two'
}, {
  id: 3,
  name: 'three'
}];

const arrayTwo = [{
  id: 2,
  name: 'two'
}, {
  id: 3,
  name: 'three'
}];
const arrayTwoId = arrayTwo.map(el => (el.id)); // extract id from arrayTwo


const result = arrayOne.filter(el => !arrayTwoId.includes(el.id));
console.log(result);

1. Extract all the ids from the arrayTwo.
2. filter those objects who do not match the array of ids of arrayTwo.