Suppose $X_1,X_2,\ldots,X_n$ are independent $N(0,\sigma^2)$ random variables.
For $i,j\in \{1,2,\ldots,n\}$, consider $$U=\frac{X_iX_j}{\sum_{i=1}^n X_i^2}$$
Provided $n>1$, we know that $U$ has a Beta distribution when $i=j$ :
$$U=\frac{X_i^2/\sigma^2}{\sum_{i=1}^n X_i^2/\sigma^2} \sim \text{Beta}\left(\frac12,\frac{n-1}2\right) \quad,\,i=1,2,\ldots,n$$
What can we say regarding the distribution of $U$ when $i\ne j$? What are the moments of $U$ in this case?
For $n=2$, if we transform $(X_1,X_2)$ to polar coordinates $(R,\Theta)$, then
$$U=\frac{X_1X_2}{X_1^2+X_2^2}=\frac{R^2\cos\Theta\sin\Theta}{R^2}=\frac{\sin(2\Theta)}{2}$$
Since $\Theta$ is uniformly distributed on $(0,2\pi)$, it seems $\sin(2\Theta)$ has an $\text{Arcsine}(-1,1)$ distribution with pdf
$$f(x)=\frac1{\pi \sqrt{1-x^2}}\mathbf 1_{(-1,1)}(x)$$
So that $U$ has pdf
$$f_U(u)=2 f(2u)=\frac2{\pi\sqrt{1-4u^2}}\mathbf1_{\left(-\frac12,\frac12\right)}(u)$$
If $\boldsymbol X=(X_1,X_2,\ldots,X_n)^T$, we can think of $U$ as the product of $i$th and $j$th components of the vector $\frac{\boldsymbol X}{\lVert \boldsymbol X \rVert}$. And we know that $\frac{\boldsymbol X}{\lVert \boldsymbol X \rVert}$ is uniformly distributed on the surface of a unit sphere. I am not sure if this helps in any way.