First, we will use complex analysis to evaluate the integral of interest. To that end we write for $n>m$
$$\begin{align}
\int_0^\infty\frac{x^{2m}}{1+x^{2n}}\,dx&=\frac12\oint_C\frac{z^{2m}}{1+z^{2n}}\,dx\\\\
&=\pi i \sum_p\text{Res}\left(\frac{z^{2m}}{1+z^{2n}},z=z_p\right) \tag 1
\end{align}$$
where $C$ is the classical "infinite semi-circular contour" closed in the upper-half plane and $z_p$ are the poles of $\frac{z^{2m}}{1+z^{2n}}$ in the upper-half plane. Note that since $n>m$ the contribution from the integration over the "infinite" semi-circle is zero.
To calculate the poles in the upper-half plane, we simply find the roots of the denominator $1+z^{2n}$ in the upper-half plane, which are $z_p=e^{i(2p+1)\pi/2n}$ for $0\le p\le n-1$.
To calculate the residues, we use L'Hospital's Rule to obtain
$$\begin{align}
\text{Res}\left(\frac{z^{2m}}{1+z^{2n}},z=z_p\right)&=\lim_{z\to z_p}\left(\frac{(z-z_p)z^{2m}}{1+z^{2n}}\right)\\\\
&=\frac{z_p^{2m}}{2nz_p^{2n-1}}\\\\
&=\frac1{2n}z_p^{2m-2n+1}\\\\
&=\frac1{2n}e^{i(2p+1)(2m-2n+1)\pi/2n}\\\\
&=-\frac1{2n}e^{i(2p+1)(2m+1)\pi/2n} \tag 2
\end{align}$$
Substituting $(2)$ into $(1)$ yields
$$\begin{align}
\int_0^\infty\frac{x^{2m}}{1+x^{2n}}\,dx&=-\frac{\pi i}{2n} \sum_{p=0}^{n-1}e^{i(2p+1)(2m+1)\pi/2n}\\\\
&=-\frac{\pi i}{2n}e^{i(2m+1)\pi/2n} \sum_{p=0}^{n-1}\left(e^{i(2m+1)\pi/n}\right)^p\\\\
&=-\frac{\pi i}{2n}e^{i(2m+1)\pi/2n}\frac{2}{1-e^{i(2m+1)\pi/n}}\\\\
&=\frac{\pi}{2n\sin\left(\frac{(2m+1)\pi}{2n}\right)} \tag 3
\end{align}$$
Enforcing the substitution $1+x^{2n}\to 1/x$ yields
$$\begin{align}
\int_0^\infty\frac{x^{2m}}{1+x^{2n}}\,dx&=\frac1{2n} \int_0^1 x^{-(2m+1)/n}\,(1-x)^{((2m+1)/2n)-1}\,dx\\\\
&=\frac1{2n}B\left(\frac{2m+1}{2n},1-\frac{2m+1}{2n}\right)\\\\
&=\frac1{2n}\Gamma\left(\frac{2m+1}{2n}\right)\,\Gamma\left(1-\frac{2m+1}{2n}\right) \tag 4\\\\
\end{align}$$
Comparing $(3)$ and $(4)$ we find that
$$\Gamma\left(\frac{2m+1}{2n}\right)\,\Gamma\left(1-\frac{2m+1}{2n}\right)=\frac{\pi}{\sin\left(\frac{(2m+1)\pi}{2n}\right)} $$
which using the density of the rational numbers and then extending by analytic continuation becomes
$$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$$
as was to be shown!
Best Answer
We can prove the identity in a different way using Euler's infinite product representation of the gamma function,
$$\Gamma(x) = \lim_{n \to \infty}\Gamma_n(x) = \lim_{n \to \infty} \frac{n!n^x}{(x+1)(x+2) \cdots (x+n)}$$
The beta function defined as $\displaystyle B(x,y) = \int_0^1 t^{x-1}(1-t)^{y-1} \, dt$ satisfies the recurrence relation
$$ \tag{1}B(x,y) = \frac{x+y}{y}B(x,y+1)$$
The proof of (1) involves integration by parts and can be furnished if desired. Applying (1) $n$ times we get
$$\tag{2}B(x,y) = \frac{(x+y)(x+y+1) \cdots (x + y + n)}{y(y+1) \cdots (y +n)}B(x,y+n+1) \\ =\frac{\Gamma_n(y)}{\Gamma_n(x+y)}n^{x}B(x,y+n+1)$$
Note that
$$n^{x}B(x,y+n+1) = n^{x}\int_0^1t^{x-1}(1-t)^{y+n} \, dt,$$
and after changing variables with $s = nt$ we get
$$\tag{3}n^{x}B(x,y+n+1) = \int_0^ns^{x-1}\left(1-\frac{s}{n}\right)^{y+n} \, ds = \int_0^ns^{x-1}\left(1-\frac{s}{n}\right)^{n} \left(1-\frac{s}{n}\right)^{y}\, ds $$
Substituting into (2) using (3) yields
$$\tag{4}B(x,y) = \frac{\Gamma_n(y)}{\Gamma_n(x+y)}\int_0^ns^{x-1}\left(1-\frac{s}{n}\right)^{n} \left(1-\frac{s}{n}\right)^{y}\, ds$$
By the dominated convergence theorem,
$$\lim_{n \to \infty}\int_0^ns^{x-1}\left(1-\frac{s}{n}\right)^{n} \left(1-\frac{s}{n}\right)^{y}\, ds = \lim_{n \to \infty}\int_0^{\infty}s^{x-1}\left(1-\frac{s}{n}\right)^{n} \left(1-\frac{s}{n}\right)^{y}1_{[0,n]}\, ds \\ = \int_0^\infty s^{x-1}e^{-s}\, ds ,$$
and, since $\lim_{n \to \infty} \Gamma_n(z) = \Gamma(z)$, upon taking the limit of both sides of (4) as $n \to \infty$ we obtain
$$B(x,y) = \frac{\Gamma(y)}{\Gamma(x+y)}\int_0^\infty s^{x-1}e^{-s}\, ds = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}$$