Certainly, I've reviewed the code and found a potential issue. The problem lies in how the threads are managed and how they access shared resources. Specifically, the `printf` statements in the `work` and `work2` functions are not synchronized, leading to unexpected outputs.

To fix this, we can introduce a mutex to ensure that only one thread can access the standard output at a time. Here's the modified code:

```c
// ... (previous code remains the same)

pthread_mutex_t print_lock = PTHREAD_MUTEX_INITIALIZER;

void *work(void *arg) {
    for(int i = 0; i < 100; i++){
        pthread_mutex_lock(&print_lock); // Lock before printing
        printf("hello world %d\n", i);
        pthread_mutex_unlock(&print_lock); // Unlock after printing
    }
    return NULL;
}

void *work2(void *arg) {
    for(int i = 0; i < 10; i++){
        pthread_mutex_lock(&print_lock); // Lock before printing
        printf("ADS %d\n", i);
        pthread_mutex_unlock(&print_lock); // Unlock after printing
    }
    return (void *) 10;
}

int main(int argc, char *argv[]) {
    // ... (previous code remains the same)
    pthread_mutex_init(&print_lock, NULL);

    mythread_t t, t2;
    mythread_create(&t, work, NULL);
    mythread_create(&t2, work2, NULL);
    mythread_detach(&t);
    int *res = malloc(sizeof(int));
    mythread_join(&t2, (void **) res);
    mythread_exit();
    pthread_mutex_destroy(&print_lock);
    return 0;
}
```

In this code, I've introduced a `pthread_mutex_t` named `print_lock`. Before printing any message, a thread will lock the mutex, and after printing, it will unlock it. This ensures that only one thread is printing at a time, preventing output interference.

Please try running this code and let me know if it resolves the issue.