= Solution
Since $\dot u(t)=a_t^2>0$, invert the image clock before exit and put $\widehat\xi_u=W_{t(u)}$. Its <quadratic variation> is $6u$ and its initial value is zero, because $\Phi(0)=0$. By the <Lévy characterization of Brownian motion>, or the localized <Dambis-Dubins-Schwarz theorem>,
$$
\widehat\xi_u=\sqrt6\,\widehat B_u
$$
for a standard <Brownian motion> up to the image exit time. If that lifetime is finite, the stopped <Brownian motion> can be extended beyond it; only the stopped law is used here.
In this clock the image <mapping-out functions> obey the ordinary <Chordal Loewner equation> with driver $\sqrt6\,\widehat B_u$. Their generated trace therefore has the law of <SLE> with parameter six, stopped when it leaves $N^*$. Meanwhile the original image segment is exactly $\Phi(\gamma)$ up to exit from $N$. The clock change is increasing, so it disappears after passing to $[\gamma]$. This proves \b[the local stopped-curve equality required in part (a)].
For completeness, the original law is scale invariant: the maps $r^{-1}g_{r^2t}(rz)$ have driver $r^{-1}\xi_{r^2t}$, whose law is again $\sqrt6 B_t$ by <Brownian scaling>. Hence $(r^{-1}\gamma_{r^2t})_{t\geq0}$ has the original trace law. Together with the stopped conformal-change argument, this establishes
$$
\boxed{[\gamma]\text{ for SLE}(6)\text{ has the locality property}.}
$$
The conclusion concerns local unparameterized segments up to exit; the capacity parameter itself changes according to $du=h_t'(\xi_t)^2dt$.
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