Javascript: Make sure to understand the question properly. I misunderstood it the first time.

var sum_hourglass_array = [];
var sum;

//k is vertical
//j is horizontal

for (let k=0; k<6; k++){
    let row = arr[k];
    for(let j=0; j<6;j++){
        if ((j+2) <=5 && (k+2)<=5){
            sum = arr[k][j] + arr[k][j+1] + arr[k][j+2] + arr[k+1][j+1] + arr[k+2][j] + arr[k+2][j+1] + arr[k+2][j+2];
            sum_hourglass_array.push(sum);
        }
    } 
}
console.log(Math.max.apply(Math,sum_hourglass_array));

C#

        List<List<int>> arr = new List<List<int>>();
        int max = -63; // minValue for this problem/scope
        
        for (int i = 0; i < 6; i++)
        {
            arr.Add(Console.ReadLine().TrimEnd().Split(' ').ToList().Select(arrTemp => Convert.ToInt32(arrTemp)).ToList());
        }
        
        for(int row = 0; row < 4; row++)
        {
            for(int col = 0; col < 4; col++)
            {
                int sum = 0;
                
                sum =   arr[row][col] + arr[row][col+1] + arr[row][col+2];
                sum +=  arr[row+1][col+1];
                sum +=  arr[row+2][col] + arr[row+2][col+1] + arr[row+2][col+2];
                
                if(sum > max) max = sum;
            }
        }
        
        Console.WriteLine(max);

C

int main() {

int** arr = malloc(6 * sizeof(int*));

for (int i = 0; i < 6; i++) {
    *(arr + i) = malloc(6 * (sizeof(int)));

    char** arr_item_temp = split_string(rtrim(readline()));

    for (int j = 0; j < 6; j++) {
        int arr_item = parse_int(*(arr_item_temp + j));

        *(*(arr + i) + j) = arr_item;
    }
}

int countk = 0;
int countl = 0;
int sum = 0;
int sum_max = -63;
int k = 0;
int l = 0;

for (int i = 0; i < 4; i++) {

    for (int j = 0; j < 4; j++) {
            countk = 0;
            for (k=i; k < i+3; k++) {
                countl = 0;
                countk++;
                for (l=j; l < j+3; l++) {
                    countl++;
                    if(countk==2){
                        if (countl==2) {
                        sum+= *(*(arr + k)+l);
                        }
                    }else {
                        sum+= *(*(arr + k)+l);
                    }         
                }//end for l
            }//end for k
        if(sum>sum_max){
            sum_max=sum;
        }
        sum=0;    
    }//end for j
}//end for i 

printf("%i",sum_max);

return 0;

}

🐍 If you want to solve in your favorite IDE - you can copy """Sample Input""" there and make 2D array from it

sample_input = """0 -4 -6 0 -7 -6 -1 -2 -6 -8 -3 -1 -8 -4 -2 -8 -8 -6 -3 -1 -2 -5 -7 -4 -3 -5 -3 -6 -6 -6 -3 -6 0 -8 -6 -7"""

arr = [list(map(int, line.split())) for line in inpuT.split("\n")]

FOR JAVA

public class Solution { public static void main(String[] args) throws IOException { BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));

    List<List<Integer>> arr = new ArrayList<>();

    IntStream.range(0, 6).forEach(i -> {
        try {
            arr.add(
                Stream.of(bufferedReader.readLine().replaceAll("\\s+$", "").split(" "))
                    .map(Integer::parseInt)
                    .collect(toList())
            );
        } catch (IOException ex) {
            throw new RuntimeException(ex);
        }
    });

    int answer = Integer.MIN_VALUE;
     for (int i = 0; i <= 3; i++) { 
        for (int j = 0; j <= 3; j++) { 
            int sum = 0;

                sum += arr.get(i).get(j);     
                sum += arr.get(i).get(j + 1); 
                sum += arr.get(i).get(j + 2); 

                sum += arr.get(i + 1).get(j + 1);

                sum += arr.get(i + 2).get(j);     
                sum += arr.get(i + 2).get(j + 1); 
                sum += arr.get(i + 2).get(j + 2); 

                answer = Math.max(sum, answer);
        }
    }
    System.out.println(answer);

    bufferedReader.close();
}

}