Let $K=F_q$ and $F=F_{q^3}$, define the set A={$x \in F$ : $Tr_{F/K} (x)=0$}. Is it true that for every $x \in A$ there are $y,z \in A$ such that $x=yz$?

In characteristic $3$, yes. Proof: Let $y=1$. In characteristic $2$, yes. In all other cases, no. Proof: The symmetric bilinear form $Tr(yz)$ on the vector space $F$ is nondegenerate (for any finite separable field extension). Its restriction to the codimension one subspace $A$ is also nondegenerate, since the orthogonal complement of $A$ is spanned by $1$, which is not in $A$. Count the nonzero solutions of $x=yz$ with $x,y,z$ all in $A$: the number is $(q^21)(q1)$, because $y$ can be any nonzero vector in $A$ and $z$ can be any nonzero vector in the (onedimensional) orthogonal complement of $Ky$ in $A$. If we identify the solutions $(x,y,z)$ and $(x,cy,c^{1}z)$ for $c\in F$ then the count becomes $q^21$. This set is mapped to another set $A0$ of the same size by $(x,y,z)\mapsto x$. To decide whether the map is surjective, look at whether it's injective. In characteristic different from $2$, nondegeneracy of the form implies that there is a solution $(x,y,z)$ with $y$ and $z$ linearly independent, so that $(x,z,y)$ is an inequivalent solution. In characteristic $2$, since $Tr(y^2)$ is universally congruent to $Tr(y)^2$ mod $2$, $A$ is closed under squaring; thus every solution has the form $(x,y,ay)$ with $a\in K$. And injectivity holds because if $(x,y_1,a_1y_1)$ and $(x,y_2,a_2y_2)$ are solutions then $\frac{y_1}{y_2}$ is a square root of $\frac{a_2}{a_1}$, so an element of $K$. 

