$$(xy'+xy)+xz'$$
Using boolean algebra I achieve $x + xz'$, which is pretty obvious by just looking at the problem, however I can't find another way to go after there in order to cancel the $z',$ since there is no $xz$ or similar term to cancel the product.
The below image shows the properties I am allowed to use, highlighted in blue. The properties that are not highlighted, like $1 = a + 1$, or so I am not allowed to use.
EDIT: reloaded the website and I think there was a problem with the website, I got an email from the book company and I was given a new question, since this one seemed misleading, nonetheless I appreciate those who took the time to read the question and gave their input.
EDIT $$\begin{aligned} (xy'+xy)+xz'&=x(y'+y)+xz'\qquad &\text{distributivity}\\&=x\cdot 1+xz' \qquad &\text{complement}\\ &=x(1+z') \qquad & \text{distributivity}\\&=x(z'+1) \qquad & \text{commutativity}\\ &=x\cdot 1 \qquad &\text{null element}\\&=x \qquad &\text{identity} \end{aligned}$$