Let $(R,\mathfrak m)$ be a Noetherian local and $I$ be an ideal of $R$ such that $\sqrt{I}=\mathfrak m$. Let $x\in I\setminus I \mathfrak m$ be such that $x$ is a non-zerodivisor on $R$. If $I/I^2$ is $R/I$-free, then is $\dfrac{I}{I^2+xR}$ also $R/I$-free?
(Here note that $\dfrac{I}{I^2+xR}$ is square of the image of the ideal $I$ in $R/xR$).
Counterexample: let $k$ be a field, $A=k[[t]], B=k[[t^3, t^7, t^8]], \mathfrak{m}=t^3B+t^7B+t^8B, I= t^6A\subset B, x=t^{11}.$
$(B,\mathfrak{m})$ is a noetherian local ring and $\sqrt{I}=\mathfrak{m}$ in $B$. $\bar{B}:=B/I\simeq k\oplus kt^3$ and $I/I^2=\bar{B}t^6\oplus \bar{B}t^7\oplus \bar{B}t^8$.
On the other hand, $I/I^2+xB\simeq kt^6\oplus \dots \oplus kt^{10}$ as $k$-vector spaces, so that $I/I^2+xB$ has odd length. However, any finite free $B/I$-module must have even length.