Suppose $X$ is a non-reflexive Banach space, $Z$ a closed subspace of $X^\*$, and $f$ a bounded functional on $Z$ with the property that there exists non-zero $x\in X$ such that $f(z^\*)=z^\*(x)$ for all $z^\*\in Z$. Can $f$ be always extended to a functional on $X^\*$ having the same property? If the answer is negative in general as I suspect, is there any characterization of the subspaces $Z$ (or/and functionals $f$) for which this would be true? 



Extending functionals on $X^*$
I do not understand your question: it seems that $x$ considered as a functional on $X^\*$ will be the desired extension.


EDIT (in connection with the comment of Markus below): If you look for a norm-preserving extension, then the answer is different. For all nonquasireflexive Banach spaces $X$ there exist total nonnorming subspaces $Z$ in $X^\*$. Total means that they are weak$^\*$ dense in $X^\*$. Nonnorming means that the function $\sup\{|z^\*(x)|:z^\*\in B(Z)\}$ takes arbitrarily small positive values of the unit sphere of $X$. This implies that there exist vectors in $X$ whose norms, as functionals on $Z$, can be very small, but the unique weak$^\*$ continuous (by weak$^\*$ density) extensions to $X^\*$ have norm $1$. 


See my [survey](https://arxiv.org/abs/math/0203139) for suitable references. I think that checking this literature one can answer majority of questions in this direction, at least in the separable case. For example, one can show that there is a norm-preserving extension for each $x$ if and only if $Z$ is $1$-norming.


