There are several ways to do this, lets see them one by one:

1. delete method: The most common way

const myObject = {
    "employeeid": "160915848",
    "firstName": "tet",
    "lastName": "test",
    "email": "test@email.com",
    "country": "Brasil",
    "currentIndustry": "aaaaaaaaaaaaa",
    "otherIndustry": "aaaaaaaaaaaaa",
    "currentOrganization": "test",
    "salary": "1234567"
};

delete myObject['currentIndustry'];
// OR delete myObject.currentIndustry;
  
console.log(myObject);

2. By making key value undefined: Alternate & a faster way:

let myObject = {
    "employeeid": "160915848",
    "firstName": "tet",
    "lastName": "test",
    "email": "test@email.com",
    "country": "Brasil",
    "currentIndustry": "aaaaaaaaaaaaa",
    "otherIndustry": "aaaaaaaaaaaaa",
    "currentOrganization": "test",
    "salary": "1234567"
  };

myObject.currentIndustry = undefined;
myObject = JSON.parse(JSON.stringify(myObject));

console.log(myObject);

3. With es6 spread Operator:

const myObject = {
    "employeeid": "160915848",
    "firstName": "tet",
    "lastName": "test",
    "email": "test@email.com",
    "country": "Brasil",
    "currentIndustry": "aaaaaaaaaaaaa",
    "otherIndustry": "aaaaaaaaaaaaa",
    "currentOrganization": "test",
    "salary": "1234567"
};


const {currentIndustry, ...filteredObject} = myObject;
console.log(filteredObject);

Or if you can use omit() of underscore js library:

const filteredObject = _.omit(currentIndustry, 'myObject');
console.log(filteredObject);

When to use what??

If you don't wanna create a new filtered object, simply go for either option 1 or 2. Make sure you define your object with let while going with the second option as we are overriding the values. Or else you can use any of them.

hope this helps :)

Another solution that avoids mutating the original array and uses a more modern features (object rest operator & destructuring)

const arr = [
    {
        XXX: "2",
        YYY: "3",
        ZZZ: "4"
    },
    {
        XXX: "5",
        YYY: "6",
        ZZZ: "7"
    },
    {
        XXX: "1",
        YYY: "2",
        ZZZ: "3"
    }
]

// destructure 'YYY' and return the other props only
const newArray = arr.map(({YYY, ...rest}) => rest)

console.log(newArray)