Assuming standard results on lifting idempotents, it's not hard to check that a path algebra $kQ$ satisfies the lifting property that Ginzburg uses to define formal smoothness in Definition 19.1.1.

If $B$ is an algebra with a nilpotent ideal $I$, we need to check that every homomorphism $\varphi:kQ\to B/I$ lifts to a homomorphism $\tilde{\varphi}:kQ\to B$.

Let $\{e_1,\dots,e_n\}$ be the idempotents in $kQ$ corresponding to the vertices of $Q$. Then $\{\varphi(e_1),\dots,\varphi(e_n)\}$ is a set of orthogonal idempotents in $B/I$ whose sum is $1$.

These lift to a set $\{f_1,\dots,f_n\}$ of orthogonal idempotents in $B$ whose sum is $1$.

For each arrow $a_s$ of $Q$ from vertex $i$ to vertex $j$, pick an arbitrary lift $\widetilde{\varphi(a_s)}$ of $\varphi(a_s)$, and let $b_s=f_i\widetilde{\varphi(a_s)}f_j$, so $b_s$ is also a lift of $\varphi(a_s)$.

Then there is a unique map $\tilde{\varphi}:kQ\to B$ with $\tilde{\varphi}(e_i)=f_i$ and $\tilde{\varphi}(a_s)=b_s$ for all $i$ and $s$, and this is a lift of $\varphi$.

This doesn't work for quivers with relations. For example, if $Q$ is an acyclic quiver and $A=kQ/I$ with $I$ a nonzero admissible ideal, then the identity map $A\to kQ/I$ does not lift to a homomorphism $A\to kQ$.