Given that,
Let’s check all of the options.
$A. \; (x^{2} – z^{2})$ has to be positive.
Let’s take $x=-1, z=2$
Now, $(-1)^{2} – (2)^{2}>0$
$\Rightarrow 1-4>0$
$\Rightarrow \boxed{-3>0 \; (False)}$
$B. \; yz can be less than one.
$yz<1$
$\Rightarrow \frac{1}{4} \times 2<1$
$\Rightarrow \boxed{\frac{1}{2}<1 \; (True)}$
$C: xy$ can never be zero.
$\Rightarrow xy \ne 0$
$\Rightarrow \boxed{xy \ne 0} (True)$
$D. \; (y^{2} – z^{2})$ is always negative.
$\Rightarrow y^{2} – z^{2} <0$
Here $y<z$
$\Rightarrow y^{2} < z^{2}$
$\Rightarrow \boxed{y^{2} - z^{2} < 0 (True)Always}$
Correct Answer $:\text{A}$