A humble Chi sqare test is probably all you need to test the null hypothesis that there is no relationship between mother tongue and city ie that the proportion of speakers is the same in each city (note that this is not the same as all proportions are a third).
As per my comments, I'm not sure this is that useful a question, depending on the context. After all, you would expect different cities to have different proportions of languages wouldn't you, on historical, geographical and cultural grounds? So you will almost certainly reject a null hypothesis of equal proportions.
But the test would be something like the below. The numbers in the table represent the number from a sample reporting that language as their mother tongue (made-up data).
> x <- data.frame(
+ row.names=c("London", "New York", "Hanover"),
+ english=c(100,100,10),
+ german=c(5,8,60),
+ french=c(7,4,12))
> x
english german french
London 100 5 7
New York 100 8 4
Hanover 10 60 12
>
> # inbuilt chi square test:
> chisq.test(x)
Pearson's Chi-squared test
data: x
X-squared = 174.4, df = 4, p-value < 2.2e-16
>
> # or, by hand:
> # First, what are the "expected" values if there
> # is no relationship between city and language
> e <- apply(x,1,sum) %o% apply(x,2,sum)/sum(x)
> e
english german french
London 76.86 26.72 8.418
New York 76.86 26.72 8.418
Hanover 56.27 19.56 6.163
> sum((x-e)^2/e)
[1] 174.4
>