It's a bit difficult to tell what they are looking for, they may just want you to go like this:

$\displaystyle x(y-z)(y+z) + y(z-x)(z+x) + z(x-y)(x+y)$

This works because:

Lets say you start with the expression

$\displaystyle a^2 - b^2$

you can factor it like this: add and subtract ab (note that ab - ab = 0 so we are adding the equivalent of zero, and thus not changing the problem)

$\displaystyle = a^2 + ab -ab -b^2$

group them together

$\displaystyle = (a^2 + ab) + (-ab -b^2)$

factor "a" out of the first group, "-b" out of the second group

$\displaystyle = a(a + b) - b(a +b)$

factor (a+b) out of each group

$\displaystyle = (a-b)(a + b)$

For that last step, if it was hard to see, replace "(a+b)" with some other letter, like "c" it would then look like this:

$\displaystyle ac - bc = c(a-b)$

Then you can replace c with "(a+b)" again, and see that it looks proper.

So from this, we can see that any time we have any terms a^2 and b^2 we can directly factor it to this:

$\displaystyle a^2 - b^2 = (a+b)(a-b)$

Be careful, though, if they have a plus sign instead of a minus sign, it does not work.

So for your problem, the first group is

$\displaystyle x(y^2 - z^2)$

$\displaystyle =x(y^2 +yz - yz - z^2)$

$\displaystyle =x(y(y +z) -z(y + z))$

$\displaystyle =x(y -z)(y + z)$

Then you can subsequently do that for the rest of the terms.

...but the instructions are somewhat ambiguous, this may not be what they are looking for. They may be looking for what mathceleb posted, it is difficult to say.