I think your final answer should be x'y + xz + yz, if you introduce xx' which evaluates to zero, you will be able to get to the final answer. Nevertheless, please find the complete solution:
(x + y)(x' + z)(y + z)
(xx' + yx' + zx + zy)(y + z)
(x'y + xyz + yz + x'yz + xz + yz)
(x'y + yz + xz)
introduce xx'
(x' + z)(y + x)
Expanding the above:
x'y + yz + x'x + xz => x'y + yz + xz which is equal to above
Hope this has answered your question
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