How to show that $\sum\limits_{k=0}^n (-1)^k\tfrac{{ {n}\choose{k}}}{{ {x+k}\choose{k}}} = \frac{x}{x+n}$
We recall the Melzak's identity $$f\left(x+y\right)=x\dbinom{x+n}{n}\sum_{k=0}^{n}\left(-1\right)^{k}\dbinom{n}{k}\frac{f\left(y-k\right)}{x+k},\, x,y\in\mathbb{R},\, x\neq-k $$ where $f $ is an algebraic polynomial up to degree $n $. So taking $f\left(z\right)\equiv1 $ we get $$\frac{1}{x\dbinom{x+n}{n}}=\sum_{k=0}^{n}\left(-1\right)^{k}\dbinom{n}{k}\frac{1}{x+k} $$ hence, by the binomial inversion, we have $$\sum_{k=0}^{n}\left(-1\right)^{k}\dbinom{n}{k}\frac{1}{\dbinom{x+k}{k}}=\color{red}{\frac{x}{x+n}}.$$
Here is a variation based upon telescoping.
We obtain \begin{align*} \sum_{k=0}^n&(-1)^k\frac{\binom{n}{k}}{\binom{x+k}{k}}\\ &=\frac{1}{\binom{x+n}{n}}\sum_{k=0}^n(-1)^k\binom{x+n}{n-k}\tag{1}\\ &=\frac{1}{\binom{x+n}{n}}\left[\sum_{k=0}^{n-1}(-1)^k\left(\binom{x+n-1}{n-k}+\binom{x+n-1}{n-k-1}\right) +(-1)^n\right]\tag{2}\\ &=\frac{1}{\binom{x+n}{n}}\left[\sum_{k=0}^{n-1}(-1)^k\binom{x+n-1}{n-k}-\sum_{k=1}^n(-1)^{k}\binom{x+n-1}{n-k}+(-1)^n\right]\tag{3}\\ &=\frac{\binom{x+n-1}{n}}{\binom{x+n}{n}}\tag{4}\\ &=\frac{x}{x+n} \end{align*} and the claim follows.
Comment:
In (1) we use $ \binom{n}{k}\binom{x+k}{k}^{-1}=\binom{x+n}{n}^{-1}\binom{x+n}{n-k} $
In (2) we use the binomial identity $\binom{p}{q}=\binom{p-1}{q}+\binom{p-1}{q-1}$
In (3) we shift the index of the right-hand series by one
In (4) we apply the telescoping
A well-known identity related to the Beta function is $$\int_0^1 t^{p-1} (1-t)^{q-1} \; dt = \frac{\Gamma(p) \Gamma(q)}{\Gamma(p+q)}$$ from which we can easily derive $$\frac{1}{\binom{n}{k}} = (n+1)\int_0^1 t^k (1-t)^{n-k} \;dt$$ so $$\sum_{k=0}^n (-1)^k \binom{n}{k} \frac{1}{\binom{x+k}{k}} = \sum_{k=0}^n (-1)^k \binom{n}{k} (x+k+1) \int_0^1 t^k (1-t)^x \;dt = I_1+I_2$$
where we define $$I_1 = \int_0^1 (1-t)^x x \sum_{k=0}^n (-1)^k \binom{n}{k} t^k \; dt$$ and $$I_2 = \int_0^1 (1-t)^x \sum_{k=0}^n (-1)^k \binom{n}{k} (k+1) t^k \; dt$$ In $I_1$, substitute $\sum_{k=0}^n (-1)^k \binom{n}{k} t^k = (1-t)^n$, with result $$I_1 = \int_0^1 x (1-t)^{n+x} \;dt = \frac{x}{(n+x)(1+n+x)}$$
Differentiating $t (1-t)^n =\sum_{k=0}^n (-1)^k \binom{n}{k} t^{k+1}$ with respect to $t$, we find $(1-t)^{n-1} [1 - (n+1)t] = \sum_{k=0}^n (-1)^k \binom{n}{k} (k+1) t^k$. Substituting into $I_2$, $$I_2 = \int_0^1 (1-t)^{x+n-1}[1-(n+1)t] \;dt =\frac{x}{1+n+x}$$ Combining these results and simplifying, we have $$I_1+I_2 = \frac{x}{x+n}$$ which is the desired result.