In chapter 3 of Vaughan's book on the Hardy–Littlewood method, he states the identity $$\sum_{X< y\leqslant n} \Lambda(y)e(\alpha y) + \sum_{X< x \leqslant n}\sum_{X<y\leqslant n/x}\sum_{\substack{d\mid x\\ d\leqslant X}}\mu(d)\,\Lambda(y)\,e(\alpha xy) = \sum_{d\leqslant X}\sum_{X<y\leqslant n/d}\sum_{z\leqslant n/(yd)}\mu(d)\,\Lambda(y)\,e(\alpha dyz),$$ where $\Lambda$ is the von Mangoldt function, $\mu$ is the Möbius function, and $e({}\cdot{}) = e^{2\pi i(\,\cdot\,)}$.
The justification given for this is the fact that $\sum_{d\mid x, d\leqslant X}\mu(d) = 1$ for $x=1$ and $0$ for $1<x\leqslant X$, but I can't seem to manipulate the LHS to obtain the RHS. I appreciate any assistance with this.
Merging the two sums on the left-hand side (after changing the order of the $x$ and $y$ summation in the second sum) gives
$$ \sum_{X<y\leq n} \Lambda(y)\left( e(\alpha y)+\sum_{X<x\leq n/y}e(\alpha xy)\sum_{\substack{d\mid x\\ d\leq X}} \mu(d)\right). $$
Using the identity you mention, we can write
$$ e(\alpha y) = \sum_{1\leq x\leq X}e(\alpha xy)\sum_{\substack{d\mid x\\ d\leq X}}\mu(d),$$
since every term vanishes except the $x=1$ term. Therefore the left-hand side is
$$ \sum_{X<y\leq n}\Lambda(y)\sum_{1\leq x\leq n/y} e(\alpha xy)\sum_{\substack{d\mid x\\ d\leq X}}\mu(d). $$
If we write $x=dz$, say, then this is $$ \sum_{X<y\leq n}\Lambda(y)\sum_{1\leq d\leq \min(n/y,X)}\sum_{z\leq n/yd} e(\alpha dyz)\mu(d). $$
Swapping the range of summation again so that $d$ is on the outside we get the right-hand side.