If $U\sim\chi_{m}^2$ independently of $V\sim\chi_n^2$ then prove that $\frac{V}{U+V}\sim\beta\left(\frac n2,\frac m2\right)$ 
If $U\sim\chi_{m}^2,V\sim\chi_n^2$ and $U,V$ are independent then prove that $\frac{V}{U+V}\sim\beta\left(\frac n2,\frac m2\right)$ 

The joint pdf of $U$ and $V$ is,
\begin{align}
f_{UV}(u,v)&=\frac{1}{2^{\frac m2}\Gamma\left(\frac m2\right)}u^{\frac m2-1}e^{-\frac u2}\frac{1}{2^{\frac n2}\Gamma\left(\frac n2\right)}v^{\frac n2-1}e^{-\frac u2}\\
&=\frac{1}{2^{\frac{m+n}{2}}\Gamma\left(\frac m2\right)\Gamma\left(\frac n2\right)}u^{\frac m2-1}v^{\frac n2-1}e^{-\frac12(u+v)}
\end{align}
Now let $Y=\frac{V}{U+V}$ then CDF of $Y$ is,
\begin{align}
F_Y(y)&=\mathbb P(Y\le y)\\
&=\mathbb P\left(\frac{V}{U+V}\le y\right)\\
&=\mathbb P\left(\frac VU\le \left(\frac{y}{1-y}\right)\right)\\
&=\mathbb P\left(V\le \left(\frac{y}{1-y}\right)U\right)\\
&=\int_{u=0}^{\infty}\int_{v=0}^{\left(\frac{y}{1-y}\right)u}\frac{1}{2^{\frac{m+n}{2}}\Gamma\left(\frac m2\right)\Gamma\left(\frac n2\right)}u^{\frac m2-1}v^{\frac n2-1}e^{-\frac12(u+v)}\:dv\:du
\end{align}
Now we can get $f(y)$ using Leibniz integral rule,
\begin{align}
f_y(y)&=\frac{1}{2^{\frac{m+n}{2}}\Gamma\left(\frac m2\right)\Gamma\left(\frac n2\right)}\underbrace{\int_{u=0}^{\infty}\frac{u}{(1-y)^2}u^{\frac m2-1}{\left(\frac{yu}{1-y}\right)}^{\frac n2-1}e^{-\frac12\left(u+\frac{yu}{1-y}\right)}\:du}_{I}
\end{align}
But it seems I am far away to $\beta\left(\frac n2,\frac m2\right)$. Is there other way to proof it$?$ Any hint or solution will be appreciated.

Update:
[For seek of completeness]Using @NCh answer,
Replace $t=u\left(\frac{1}{2(1-y)}\right)$, $u=2(1-y)t$, $du=2(1-y)\,dt$:
$$
I=\frac{y^{\frac{n}2-1}}{(1-y)^{\frac{n}{2}+1}}\cdot 2^{\frac{n+m}{2}}(1-y)^\frac{n+m}{2}\underbrace{\int_{t=0}^{\infty}t^{\frac{n+m}{2}-1}e^{-t}\:dt}_{\Gamma\left(\frac{n+m}{2}\right)}$$
$$f_Y(y)=\frac{\Gamma\left(\frac{n+m}{2}\right)}{\Gamma\left(\frac m2\right)\Gamma\left(\frac n2\right)}y^{\frac{n}2-1}(1-y)^{\frac m2-1}$$
Hence $f_Y(y)\sim \beta\left(\frac n2,\frac m2\right)$
 A: In addition to brilliant comment of StubbornAtom, I’ll only give a partial answer on how to get to the beta distribution in your solution.
First, replace $\Gamma\left(\frac12\right)$ in denominators by $\Gamma\left(\frac{n}2\right)$ and $\Gamma\left(\frac{m}2\right)$ respectively to correct misprints in your solution. Then consider the integral
$$
\int_{u=0}^{\infty}\frac{u}{(1-y)^2}u^{\frac m2-1}{\left(\frac{yu}{1-y}\right)}^{\frac n2-1}e^{-\frac12\left(u+\frac{yu}{1-y}\right)}\:du
$$
$$ = 
\frac{y^{\frac{n}2-1}}{(1-y)^{\frac{n}{2}+1}}\int_{u=0}^{\infty}u^{\frac{n+m}{2}-1}e^{-u\left(\frac{1}{2(1-y)}\right)}\:du := I
$$
Replace $t=u\left(\frac{1}{2(1-y)}\right)$, $u=2(1-y)t$, $du=2(1-y)\,dt$:
$$
I=\frac{y^{\frac{n}2-1}}{(1-y)^{\frac{n}{2}+1}}\cdot 2^{\frac{n+m}{2}}(1-y)^\frac{n+m}{2}\underbrace{\int_{t=0}^{\infty}t^{\frac{n+m}{2}-1}e^{-t}\:dt}_{\Gamma\left(\frac{n+m}{2}\right)}.
$$
Finally, substitute this value back into the pdf, you will get the desired pdf.
A: \begin{align}
\Pr\{\frac{V}{U+V}\leq y\}&=\Pr\{\frac{1}{\frac mn\frac{U/m}{V/n}+1}\leq y\}\\
&=\Pr\{\frac{1}{\frac mnF_{m,n}+1}\leq y\}\\
&=\Pr\{F_{m,n} \geq (y^{-1}-1)\frac nm\}\\
&=1-\Pr\{F_{m,n} \leq (y^{-1}-1)\frac nm\}\\
&=1-\frac{B\left(\left(y^{-1}-1\right)\frac nm,m,n\right)}{B(m,n)},\\
\end{align}
where $B\left(x,a,b\right)=\int_0^x t^{a-1}(1-t)^{b-1}dt$. See here for some more details. Therefore the pdf of $\frac{V}{U+V}$ is $$\frac d{dy}\left(1-\frac{B\left(\left(y^{-1}-1\right)\frac nm,m,n\right)}{B(m,n)}\right).$$
The rest shall be manageable. 
