Prove that $\sum_{k=0}^n (-1)^k\frac{{ {n}\choose{k}}{ {x}\choose{k}}}{{ {y}\choose{k}}} = \frac{{ {y-x}\choose{n}}}{{ {y}\choose{n}}}$

Note that

$$\binom{y}n\binom{n}k=\binom{y}k\binom{y-k}{n-k}\;,$$

so

$$\binom{y}n\cdot\frac{\binom{n}k\binom{x}k}{\binom{y}k}=\binom{y-k}{n-k}\binom{x}k\;,$$

and after multiplication by $\binom{y}n$, the proposed identity reduces to

$$\sum_{k=0}^n(-1)^k\binom{x}k\binom{y-k}{n-k}=\binom{y-x}n\;.\tag{1}$$

$\binom{y-x}n$ is the number of $n$-element subsets of $[y]\setminus[x]$, where as usual $[x]=\{1,\ldots,x\}$. For $k\in[x]$ let $\mathscr{A}_k$ be the family of $n$-element subsets of $[y]$ that contain $k$. Then

$$\left|\,\bigcap_{k\in I}\mathscr{A}_k\,\right|=\binom{y-|I|}{n-|I|}$$

whenever $\varnothing\ne I\subseteq[x]$, so by the inclusion-exclusion principle we have

$$\begin{align*} \left|\,\bigcup_{k\in[x]}\mathscr{A}_k\,\right|&=\sum_{\varnothing\ne I\subseteq[x]}(-1)^{|I|+1}\binom{y-|I|}{n-|I|}\\ &=\sum_{k=1}^x(-1)^{k+1}\binom{x}k\binom{y-k}{n-k}\\ &=\sum_{k=1}^n(-1)^{k+1}\binom{x}k\binom{y-k}{n-k}\;, \end{align*}$$

assuming that $x\ge n$. (Recall that by definition $\binom{y-k}{n-k}=0$ when $n-k<0$.) This is the number of $n$-element subsets of $[y]$ that do intersect $[x]$, so

$$\begin{align*}\binom{y-x}n&=\binom{y}n-\left|\,\bigcup_{k\in[x]}\mathscr{A}_k\,\right|\\ &=\binom{y}n-\sum_{k=1}^n(-1)^{k+1}\binom{x}k\binom{y-k}{n-k}\\ &=\binom{y}n+\sum_{k=1}^n(-1)^k\binom{x}k\binom{y-k}{n-k}\\ &=\sum_{k=0}^n(-1)^k\binom{x}k\binom{y-k}{n-k}\;, \end{align*}$$

as desired.

Added: as darij grinberg pointed out in the comments, the combinatorial interpretation requires the assumption that $y\ge x$. For each $x$ the combinatorial argument establishes $(1)$ for each integer $y\ge x$ and each side of $(1)$ is an $n$-th degree polynomial in $y$, so $(1)$ must be a polynomial identity.