$\def\st{\operatorname{st}}$ I'm studying non-standard calc from Keisler's book. Taking "standard part" rule doesn't make sense... its not commutative.
e.g.
$a$ is finite non infinitesimal
$b,c$ are infinitesimal
$\st(a\cdot{b\over c}) = \st(a) \cdot \st({b\over c}) = a \cdot \st({b\over c}) $ [$\st({b\over c})$ is indeterminate so we leave it as is]
$\st(b\cdot {a\over c}) = \st(b) \cdot ...... = 0$ [$b$ is infinitesimal so $\st(b)$ is $0$]
$2$ very different answers for the same expression.. what's the deal here ?
You are applying the property that $$st(xy)=st(x)st(y)$$ However this applies only when $x,y$ are finite; see Theorem 3 on p.37.
In your second calculation, $a/c$ is not finite. Your first calculation might be incorrect as well, depending on whether or not $b/c$ is finite.