Q:

Variance of $\frac{x_i}{\sum_{i=1}^nx_i}$ when $x_1,\ldots,x_n$ are iid random variables

Suppose $x_1,\ldots,x_n$ are iid random variables with finite variances. Is the variance of $z_i=\frac{x_i}{\sum_{i=1}^nx_i}$ a function of $n$ only? I am interested in two cases. Case 1: the variance of $x_i$ is zero. Case 2: the variance of $x_i$ is nonzero. 
I know that the expectation is $\frac{1}{n}$ because $\sum_{i=1}^nz_i=1$ so that both $\sum_{i=1}^nE(z_i)=nE(z_i)$ and $\sum_{i=1}^nE(z_i)=E(\sum_{i=1}^nz_i)=1$. To find the variance $V(z_i)$ we would want to find $E(z_i^2)$, and we could do so by finding $E(\sum_{i=1}^nz_i^2)=E(\frac{\sum_{i=1}^nx_i^2}{(\sum_{i=1}^nx_i)^2})$.

A:

As clarified in the comments, the question is whether the variance is a function of $n$ only or depends on the distribution. 
It depends on the distribution. The variance of $z_i$ is zero or not according as the variance of the $x_i$ is zero or not.
For a slightly more interesting case, consider $x_i$ uniformly distributed over $\{a,b\}$. Then $z_i$ takes the value $\frac12$ with probability $\frac12$ and the values $\frac a{a+b}$ and $\frac b{a+b}$ with probabilities $\frac14$ each, so the variance is
$$
\frac14\left(0+0+\left(\frac a{a+b}-\frac12\right)^2+\left(\frac b{a+b}-\frac12\right)^2\right)=\frac18-\frac{ab}{2(a+b)^2}\;,
$$
which depends on $a$ and $b$ and thus on the distribution.

