I have to show that this two rings are isomorphic $$(\mathbb{Q}[X,Y]/\langle XY-1\rangle)_{\langle X-1,Y-1 \rangle} \simeq \mathbb{Q}[X]_{\langle X-1\rangle}$$
I tried using the fact that $$ \mathbb{Q}[X,Y]/\langle XY-1\rangle \simeq \mathbb{Q}[X]_{X}$$ but I don't know what to do with the localization on $\langle X-1,Y-1\rangle$ since the $Y$ disappears. Can anyone give me some hints as to how to proceed?
First, why should we expect these rings to be isomorphic? You didn't mention this in your question, but it's an important part of solving these kinds of problems "in the wild".
Notice your first ring $R = \mathbb{Q}[x,y] \big / \langle xy - 1 \rangle$ is a hyperbola, and $S = \mathbb{Q}[x]$ is a line. So your question is asking why, locally, a hyperbola at $(x,y) = (1,1)$ looks the same as a line at $x = 1$. Geometrically this is clear, but we want to formalize the idea algebraically. There are actually a couple ways you might do this, but since you were working towards a solution, let's try to finish up your idea!
You're right that $R \cong \mathbb{Q}[x]_x$, so we're trying to compute $(\mathbb{Q}[x]_x)_{\langle x-1,y-1 \rangle}$. And you ask the perfectly reasonable question "what does this mean?". After all, there doesn't seem to be a $y$ in $\mathbb{Q}[x]$, so how should we proceed?
The answer is that $\langle x-1,y-1 \rangle$ doesn't make sense inside $\mathbb{Q}[x]_x$. We used an isomorphism of rings, and we need to look at $\langle x-1,y-1 \rangle$ under this isomorphism! Notably in $\mathbb{Q}[x]_x$ we think of $y$ as $\frac{1}{x}$, so our ideal is really $\langle x-1, \frac{1}{x} - 1 \rangle$. Note, despite what the commenter says, $\frac{1}{x} - 1$ is not a unit in $\mathbb{Q}[x]_x$! This is ok, though, as we have a simple approach: $\langle x-1, \frac{1}{x} - 1 \rangle = \langle x-1 \rangle$ (do you see why?). So we know
$$ R_{\langle x-1,y-1 \rangle} \cong (\mathbb{Q}[x]_x)_{\langle x-1 \rangle} $$
But what is this ring? $\mathbb{Q}[x]_x$ says "take $\mathbb{Q}[x]$ and make $x$ invertible", but then localizing at $\langle x-1 \rangle$ says "also make everything outside $\langle x-1 \rangle$ invertible". Since $x \not \in \langle x-1 \rangle$, we're doing extra work by making it invertible twice, and
$$ (\mathbb{Q}[x]_x)_{\langle x-1 \rangle} \cong \mathbb{Q}[x]_{\langle x-1 \rangle} $$
which is what we wanted.
I hope this helps ^_^