Independence of ratios of independent variates

If $$X= x_1/(x_1+x_2)$$ and $$Y= (x_1+x_2)/(x_1+x_2+x_3)$$ where $$x_1,x_2,x_3$$ independent chi-square variates with d.f $$n_1,n_2,n_3$$ respectively, are $$X$$ & $$Y$$ independent?

I know the condition for independence of two random variables and the usual method followed i.e using joint distribution . But is there any short way to establish independence in the above question?

It is a "well-known" property of the Gamma distributions that $$x_1/(x_1+x_2)$$ and $$(x_1+x_2)$$ are independent, and that $$x_1+x_2+x_3$$ and $$Y= (x_1+x_2)/(x_1+x_2+x_3)$$ are independent. For instance, writing \begin{align*} x_1&=\{\varrho \sin(\theta)\}^2\\ x_1&=\{\varrho \cos(\theta)\}^2\\ \end{align*} we get that $$X=\frac{\{\varrho \sin(\theta)\}^2}{\{\varrho \sin(\theta)\}^2+\{\varrho \cos(\theta)\}^2}=\sin(\theta)^2$$ and $$Y=\frac{\varrho^2}{\varrho^2+x_3}$$ are indeed functions of different variates (although the independence between $$\varrho$$ and $$\theta$$ has to be established).
You could view the chi-squared variables $$x_1,x_2,x_3$$ as relating to independent standard normal distributed variables which in it's turn relates to uniformly distributed variables on a n-sphere https://en.wikipedia.org/wiki/N-sphere#Generating_random_points
The distribution of $$\frac{x_1}{x_1+x_2}$$ "the point/angle on a circle" is independent from $$\frac{x_1+x_2}{x_1+x_2+x_3}$$ "the (relative) squared radius of the circle" or $$x_1+x_2+x_3$$ "the squared radius of the sphere in which that circle is embedded".
You could take a n-sphere (embedded in dimension $$n_1+n_2+n_3$$) and compute explicitly the value for $$f_X(x)$$ by computing the relative ratio of the areas of sub-sphere, the $$(n_1-1)$$-sphere with radius $$\sqrt{x_1}$$ and the $$(n_1-n_2-1)$$-sphere with radius $$\sqrt{x_1+x_2}$$. The result should only depend on the fraction $$X=\frac{x_1}{x_1+x_2}$$ and be independent from $$x_1+x_2$$ or $$x_1+x_2+x_3$$.