We can suppose that $x_i\leq x_j$ for $i (otherwise just permute the components). As I'll show below, the determinant of your matrix is $\det C_n=\frac{x_2^2-x_1^2}{x_2^2}\frac{x_3^2-x_2^2}{x_3^2}\dots\frac{x_n^2-x_{n-1}^2}{x_n^2},\qquad(*)$ which is $\geq 0$, and it is $>0$ iff all $x_i$'s are different. Notice that if you take the top left corner of $C$ with $k$ rows and columns, you get $C_k$, and $\det C_k\geq0$. Your matrix is therefore positive semidefinite (by Sylvester criterion), and it is positive definite iff all $x_i$'s are different.
Now we need to prove $(*)$. Let $D_n$ be $C_n$ with $i,k$-th element multiplied by $x_i x_j$, so that $\det D_n=\det C_n\times\prod_i x_i^2$. We want to show
$\det D_n=x_1^2(x_2^2-x_1^2)\dots(x_n^2-x_{n-1}^2).$
$D_n$ looks like $ \begin{pmatrix} x_1^2 & x_1^2& x_1^2&x_1^2\\ x_1^2& x_2^2& x_2^2&x_2^2\\ x_1^2& x_2^2&x_3^2&x_3^2\\ x_1^2& x_2^2&x_3^2&x_4^2 \end{pmatrix} $ (for $n=4$ - I hope the pattern is clear), i.e. $ \begin{pmatrix} a & a& a&a\\ a& b& b&b\\ a& b&c&c\\ a& b&c&d \end{pmatrix}. $ ($a=x_1^2,\dots,d=x_4^2$). If we now subtract the first row from the others and then the first column from the others, we get $ \begin{pmatrix} a & 0& 0&0\\ 0& b-a& b-a&b-a\\ 0& b-a&c-a&c-a\\ 0& b-a&c-a&d-a \end{pmatrix}. $ i.e. $ \det \begin{pmatrix} a & a& a&a\\ a& b& b&b\\ a& b&c&c\\ a& b&c&d \end{pmatrix}= a\det \begin{pmatrix} b-a& b-a&b-a\\ b-a&c-a&c-a\\ b-a&c-a&d-a \end{pmatrix} $ Repeating this identity, we get $\det D_n= \det \begin{pmatrix} a & a& a&a\\ a& b& b&b\\ a& b&c&c\\ a& b&c&d \end{pmatrix}= a\det \begin{pmatrix} b-a& b-a&b-a\\ b-a&c-a&c-a\\ b-a&c-a&d-a \end{pmatrix}= a(b-a)\det \begin{pmatrix} c-b&c-b\\ c-b&d-b \end{pmatrix}= a(b-a)(c-b)(d-c) $ as we wanted to show.
There must be a more intelligent solution, but for the moment this one should do.