How to find if a metric space is complete? Specifically, I need to find if the metric space $(\mathbb{R}, d)$ with the metric $$d(x,y)=\dfrac{|x-y|}{\sqrt{1+x^2}\sqrt{1+y^2}}$$ is complete or not, but I didn't get how to show if a space is complete and, in case it isn't, how to find a counterexample.

Thinking about Cauchy completeness, one observation is that $d(x,y)$ can get small not only if $x$ and $y$ are close to each other in the usual metric, but also if one or both of them are large. This is likely the area where 'weird' effects can happen, leading to investigate sequences that are unbounded. The simplest one is $(n)_{n\in \mathbb{N}}$.
The square roots in the definition are 'almost' $x$ or $y$, but just slightly modified to allow for the cases $x=0$ or $y=0$. So intuitively, $d(x,y)$ is about the same as $\frac{|x-y|}{|x|\cdot |y|}$, for large numbers $x,y$:
$$\frac{1}{2}\cdot \left| \frac{1}{x} - \frac{1}{y}\right| = \frac{|x-y|}{2 |x| \cdot |y|} \leq d(x,y)\leq \frac{|x-y|}{|x|\cdot |y|} = \left|\frac{1}{x}-\frac{1}{y}\right|$$
for $|x|,|y|\geq 1$.
Consider the sequence $(n)_{n\in \mathbb{N}}$. For $n,m\geq n_0$ we have $$d(n,m) \leq \left|\frac{1}{n}-\frac{1}{m}\right|\leq \frac{2}{n_0}$$ So it is a Cauchy sequence. 
For every  $x\in\mathbb{R}$ with $|x|\geq 1$, $$d(x,n) \geq \frac{1}{2}\left|\frac{1}{x} - \frac{1}{n}\right|,$$ which converges to $\frac{1}{2|x|}>0$ for $n\to \infty.$ 
For $x\in\mathbb{R}$ with $|x|< 1$, 
$$d(x,n)\geq \frac{|1-n|}{\sqrt{2}\cdot \sqrt{2 n^2}} = \frac{1}{2} \cdot \left|\frac{1}{n} - 1\right|,$$ which converges to $\frac{1}{2}>0$.
So the (Cauchy) sequence $(n)_{n\in \mathbb{N}}$ does not converge to any $x\in\mathbb{R}$ and the metric is not complete.
