Let $f(x,y,z) = xz-y^2+xyz$. We wish to calculate $\displaystyle \iint_{S} \nabla f \cdot \overline{n}d\sigma$ where $d\sigma$ is the area element of $S$ , and $\overline{n}$ is the outward pointing unit normal to $S$ at $(x,y,z)$. Applying the Divergence theorem, we get:
$$\displaystyle \iint_{S} \nabla f \cdot \overline{n}d\sigma = -2\iiint_{V} d\tau$$
My question is: where does the idea that this is the volume of the unit sphere come from?
Start with $\nabla f = (z+yz)\hat{i} + (-2y+xz)\hat{j}+ (x+xy)\hat{k}$.
Lets apply the divergence theorem and see what we have to work with.
$$ \iint_S \left[(z+yz)\hat{i} + (-2y+xz)\hat{j}+ (x+xy)\hat{k} \right] \cdot \bar{n}d\sigma = $$ $$ \iiint_V \nabla \cdot \left[ (z+yz)\hat{i} + (-2y+xz)\hat{j}+ (x+xy)\hat{k} \right] dV $$ $$ = \iiint_V \left[ 0\hat{i} -2 \hat{j} + 0\hat{k}\right] dV $$ $$ =-2 \iiint_V dV $$
This is the result you show. It states that the flux through your surface is -2 times the volume enclosed in the surface so you have a sink. Now the volume enclosed by the surface is whatever you want it to be. Usually it would be stated in the problem or it is a volume of interest by you.
Say we want to know the flux through the surface of a unit sphere. The volume is $\frac{4\pi}{3}$ so the flux through this surface is $-\frac{8\pi}{3}$.
In your question you never state what the surface, S, is.