The algebra is $A=kQ/I$ with the assumptions $I \subseteq J^2$ and $I/(JI+IJ) \neq 0$, where $J$ is the ideal in $kQ$ generated by the arrows.

Assume $A$ is hereditary. Then $J/I$ is a projective $A$-module and the exact sequence of A-modules $$0 \rightarrow I/IJ \rightarrow J/IJ \rightarrow J/I \rightarrow 0$$ must split. Let $s \colon J/IJ \rightarrow I/IJ$ denote a retraction.

Let $x$ be an element in $I \smallsetminus (JI+IJ)$. Since $I \subseteq J^2$, we can write $$x=p_1 q_1 + \cdots + p_n q_n$$ with $p_i,q_i \in J$ for all $1 \geq i \geq n$. Then $$s(p_1 q_1 + \cdots + p_n q_n+IJ)=p_1 s(q_1) + \cdots + p_n s(q_n) +IJ \in JI+IJ,$$ a contradiction.

Therefore $A$ is not hereditary.

The algebra is $A=kQ/I$ with the assumptions $I \subseteq J^2$ and $I/(JI+IJ) \neq 0$, where $J$ is the ideal in $kQ$ generated by the arrows.

Assume $A$ is hereditary. Then $J/I$ is a projective $A$-module and the exact sequence of $A$-modules $$0 \rightarrow I/IJ \rightarrow J/IJ \rightarrow J/I \rightarrow 0$$ must split. Let $s \colon J/IJ \rightarrow I/IJ$ denote a retraction.

Let $x$ be an element in $I \smallsetminus (JI+IJ)$. Since $I \subseteq J^2$, we can write $$x=p_1 q_1 + \cdots + p_n q_n$$ with $p_i,q_i \in J$ for all $1 \leq i \leq n$. Then $$x+IJ=s(p_1 q_1 + \cdots + p_n q_n+IJ)=p_1 s(q_1) + \cdots + p_n s(q_n) +IJ \in JI+IJ,$$ a contradiction.

Therefore $A$ is not hereditary.