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

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.