Use a percent-encoded URL:
link = 'http://finance.yahoo.com/news/caf%C3%A9s-growing-faster-than-fast-food-peers-144512056.html'
I found the above percent-encoded URL by pointing the browser at
http://finance.yahoo.com/news/cafés-growing-faster-than-fast-food-peers-144512056.html
going to the page, then copying-and-pasting the
encoded url supplied by the browser back into the text editor. However, you can generate a percent-encoded URL programmatically using:
from urllib import parse
link = 'http://finance.yahoo.com/news/cafés-growing-faster-than-fast-food-peers-144512056.html'
scheme, netloc, path, query, fragment = parse.urlsplit(link)
path = parse.quote(path)
link = parse.urlunsplit((scheme, netloc, path, query, fragment))
which yields
http://finance.yahoo.com/news/caf%C3%A9s-growing-faster-than-fast-food-peers-144512056.html