F(x,y,z) = (x + y')(x + y)(xz')
= (xx) + (xy) + (xy')+(y'y)(xz')
= [x + (xy) + (xy') + 0 ] (xz')
= [x + x(y + y') + 0] (xz')
= [x + x(1) + 0] (xz')
= [x + x + 0] (xz')
= [x + x] (xz')
= x*(xz')
= (xx)*z'
= xz'
xz' is my final answer.
Well yes, $$(x+y^\prime)(x+y)=x$$ and $$x(xz^\prime) = xz^\prime,$$ so your result is correct.