(Hopefully this time I did not mess up the indices in the QYBE :/ )

Let $\mathfrak g$ \mathfrak{sl}_2$ be the Heisenberg Lie algebra, spanned by $x$, e$, $y$ f$ and $z$ h$ with $z$ central [h,e]=2e$, $[h,f]=-2f$ and $[x,y]=z$. [e,f]=h$, as usual. Let $r=x\wedge y\in\Lambda^2\mathfrak g$, r=e\wedge f\in\Lambda^2\mathfrak{sl}_2$ and let $\delta=[\mathord-,r]:\mathfrak g\to\Lambda^2\mathfrak g$ be the inner derivation corresponding to $r$. Of couse $\delta$ is a $1$-cocycle, and one checks by hand that it is a cobracket; explicitely, $\delta(h)=0$, $\delta(e)=e\wedge h$ and $\delta(f)=f\wedge h$. Let $\gamma:\mathfrak g\otimes\mathfrak g\to S^3\mathfrak g$ be the map in question. Then $\gamma(x\otimes y)=z^3$\gamma(e\otimes f)\neq0$.

Let $\mathfrak g$ be the Heisenberg Lie algebra, spanned by $x$, $y$ and $z$ with $z$ central and $[x,y]=z$. Let $r=x\wedge y\in\Lambda^2\mathfrak g$, and let $\delta=[\mathord-,r]:\mathfrak g\to\Lambda^2\mathfrak g$ be the inner derivation corresponding to $r$. Of couse $\delta$ is a $1$-cocycle, and one checks by hand that it is a cobracket. Let $\gamma:\mathfrak g\otimes\mathfrak g\to S^3\mathfrak g$ be the map in question. Then $\gamma(x\otimes y)=z^3$.