I loaded a Javascript widget that outputs HTML, CSS, and additional Javascript.
The source of the page (test.html) is just
<!DOCTYPE html>
<html lang="en">
<head>
<title>Widget Test</title>
</head>
<body>
<div id="widget"></div>
<script type="text/javascript" src="http://widgets.some.site/render?element_id=widget_id&customer_id=999999&gallery=22222&widget_config=769792669" async="async">
</script>
</body>
</html>
The script executes and generates a widget. I can see the HTML nodes in the Element tab of Chrome DevTools. When inspecting a particular element, the inspector says that its style is located at test.html:239, but when I click on the link, it shows me the source page again.
When I load the page in Firebug, clicking on the line number sends me to an internal version of the stylesheet maintained by Firefox. Is there a way Chrome DevTools does this as well? I like Firebug's output of dynamically generated CSS, since I can copy and paste very easily.
Inspect Element. You can see the generated HTML there