In sum of terms, any term is an implicant because it implies the function. So, $xz$ is an implicant and hence C is the answer. Still, lets see the other options.
If no minimization is possible for an implicant (by removing any variable) it becomes a prime implicant.
If a prime implicant is present in any possible expression for a function, it is called an essential prime implicant. (For example in Kmap we might be able to choose among several prime implicants but for essential prime implicants there won't be a choice).

x'y' 
x'y 
xy 
xy' 
z' 
$1$ 
$1$ 

$1$ 
z 
$1$ 
$1$ 
$1$ 
$1$ 
${\begin{array}{ccccc}\hline\\
& \textbf{x'y'}& \textbf{x'y}&\bf{ xy}& \textbf{xy'} \\\hline
\textbf{z'}&1&1&&1 \\\hline \bf{z}&1&1&1&1\\ \hline
\end{array}}$
So, $f = x' + y'x + xz$
$= y' + x' + z$ (could be also derived using algebraic rules as in http://www.ee.surrey.ac.uk/Projects/Labview/boolalgebra/ )
So, the prime implicants are $x', y'$ and $z$. Being single variable ones and with no common variables, all must be essential also.
Now, a choice is false, as $y'$ is a prime implicant and hence, $y'x$ is just an implicant but not prime.
b choice  $xz$ is not a minterm. A minterm must include all variables. So, $xyz$ is a minterm so, is $xy'z$, but not $xz$.
d choice  $y'$ is a prime implicant not $y$.