I get the follow error with my lambda function: Calling the invoke API action failed with this message: Network Failure

I have set up a VPC with outbound traffic. Any help would be massively appreciated

See below for the code:

import os
import time
from datetime import datetime
from selenium import webdriver
from selenium.webdriver.chrome.options import Options

import boto3


def lambda_handler(event, context):
    options = Options()
    options.binary_location = '/opt/headless-chromium'
    options.add_argument('--headless')

    
     # Add your AWS S3 bucket name
    S3_BUCKET_NAME = "webscrapebucketxxxxxx22"

    # Initialize the S3 client
    s3 = boto3.client("s3")

    # Set the dictionary of website URLs you want to screenshot
    urls = {  "Google": "http://www.google.com"

    }
    
    # Set the path to your browser driver
    driver_path = "/opt/chromedriver"
    
 # Create a new instance of the browser driver with headless mode
    chrome_options = Options()
    chrome_options.add_argument("--headless")
    driver = webdriver.Chrome(executable_path=driver_path, chrome_options=options)
    
    # Define a function to take a screenshot and save it to a file
    def take_screenshot(url, name):
        # Navigate to the website
        driver.get(url)
        
        # Wait for 30 seconds to ensure that the page is fully loaded
        time.sleep(30)
        
        # Get the current date as a string in the format YYYY-MM-DD
        now = datetime.now().strftime("%Y-%m-%d_%H-%M-%S")
        
        # Generate the filename using the website name and the current date
        filename = f"{name}_{now}.png"
        
        # Save the screenshot to a file in the /tmp directory
        filepath = os.path.join("/tmp", filename)
        driver.save_screenshot(filepath)
        
        # Return the filepath for logging purposes
        return filepath
        
    def upload_to_s3(filepath):
        key = os.path.basename(filepath)
        s3.upload_file(filepath, S3_BUCKET_NAME, key)
        print(f"Uploaded {filepath} to s3://{S3_BUCKET_NAME}/{key}")
        
        
    # Loop through the dictionary of website URLs and take screenshots
    for name, url in urls.items():
        filepath = take_screenshot(url, name)
        upload_to_s3(filepath)
        print(f"Screenshot of {name} saved to {filepath}")