= Solution
Parametrize the circle by $z=e^{it}$, $0\leq t\leq2\pi$. Choose a continuous argument $\vartheta(t)$ of $\phi(e^{it})$, so $\phi(e^{it})=|\phi(e^{it})|e^{i\vartheta(t)}$. Its <winding number> about zero is
$$
w(\phi)=\frac{\vartheta(2\pi)-\vartheta(0)}{2\pi}\in\mathbb Z.
$$
The two facts proved below determine the requested calculation without needing differentiability or contour <integration>. Put $P(z)=(3z-2)(z-3)(2z+1)$. On $|z|=1$, $|P(z)|\geq1\cdot2\cdot1=2$, so adding one does not change its winding. The first and third factors have winding one, since their constant terms have modulus smaller than their linear terms; the second has winding zero, since its $z$ term has modulus smaller than its constant term. Nonzero constants have winding zero and $z$ has winding one. Thus
$$
\boxed{w(\phi)=1+0+1=2.}
$$
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