You can simplify your code and avoid unnecessary looping as soon as the first element breaking the rule is found.

Using a for loop

function isSpecial(arr) {
  for (let i = 0; i < arr.length; i++) {
    if (arr[i] % 2 !== i % 2) return false
  }
  return true;
}

Using Array.prototype.every

function isSpecial(arr) {
  return arr.every((item, index) => item % 2 === index % 2);
}

Your return ((arr[i % 2 == 0]) % 2 == 0) && ((arr[i % 2 !==0]) % 2 !== 0) syntax is wrong. Hope the below function meets your requirement.

Logic

• Loop through the array, check each node for "special" condition and store it in isNodeSpecial
• "special" condition means, even index has even number and odd index have odd number.
• Initialize a variable outside the loop which holds the isSpecial status of array.
• Update the isSpecial varaible as logical and of isSpecial and isNodeSpecial.
• Whenever a single node violates the condition the isSpecial is set to false and loop exits.

function isSpecialArray(arr) {
  let isSpecial = true;
  for (i = 0; i < arr.length && isSpecial; i++) {
    const isNodeSpecial = (i % 2 === 0) ? arr[i] % 2 === 0 : arr[i] % 2 === 1;
    isSpecial = isSpecial && isNodeSpecial;
  }
  return isSpecial;
}

console.log(isSpecialArray([2, 7, 4, 9, 6, 1, 6, 3])); // true
console.log(isSpecialArray([2, 7, 4, 10, 6, 1, 6, 3]));// false

You can check if any element does not satisfy the condition and immediately return false. If all elements satisfy the condition, then return true. That is best solution when it comes to performance. You can change your code like this:

function isSpecialArray(arr) {
  for (i = 0; i < arr.length; i++) {
    if (arr[i] % 2 !== i % 2) return false;
  }
  return true;
}

console.log(isSpecialArray([2, 7, 4, 9, 6, 1, 6, 3])) //true
console.log(isSpecialArray([1, 7, 4, 9, 6, 1, 6, 3])) //false