$f(x,y,z) = xy'z' + y'z+xy$
If we take $y'$ is common,then we get
$f(x,y,z) = y'(xz' + z)+xy$
Apply the distributive property $x+yz = (x+y)\cdot(x+z)$
$f(x,y,z) = y'\big[(x + z)\cdot(z'+z)\big]+xy$
$f(x,y,z) = y'(x + z)+xy$
$\boxed{f(x,y,z) = xy' + y'z + xy}$
$f(x,y,z) = x(y'+y) + y'z$
$\boxed{f(x,y,z) = x + y'z}$
This is minimal SOP.
$$(OR)$$
$f(x,y,z) = xy'z' + y'z+xy$
If we take $x$ is common,then we get
$f(x,y,z) = x(y'z' + y) + y'z$
Apply the distributive property $x+yz = (x+y)\cdot(x+z)$
$f(x,y,z) = x\big[(y' + y)\cdot(z'+y)\big] + y'z$
$f(x,y,z) = x(z'+y) + y'z$
$\boxed{f(x,y,z) = xz' + xy + y'z}$
$f(x,y,z)= xz' + (xy+y')\cdot(xy + z)$
$f(x,y,z)= xz' + [(x+y')\cdot(y+y')]\cdot(xy + z)$
$f(x,y,z)= xz' + (x+y')\cdot(xy + z)$
$f(x,y,z)= xz' + xyz + xz + y'z$
$f(x,y,z)= xz' + xz(y + 1) + y'z$
$f(x,y,z)= xz' + xz + y'z$
$f(x,y,z)= x(z' + z) + y'z$
$\boxed{f(x,y,z)= x+ y'z}$
This is minimal SOP.
So, both are same.
$$(OR)$$
$f(x,y,z) = xy'z' + y'z+xy$
We can expand this expression and get the canonical SOP.
$f(x,y,z) = \underbrace{xy'z'}_{100} + \underbrace{x'y'z}_{001} + \underbrace{xy'z}_{101}+\underbrace{xyz'}_{110}+ \underbrace{xyz}_{111}$
$f(x,y,z) = \sum(1,4,5,6,7)$

$\boxed{f(x,y,z)= x+ y'z}$
see this for reference: minimization