= Solution
Let $J_t=\psi_t'(U_t)$, $q_t=\psi_t''(U_t)/\psi_t'(U_t)$, and $r_t=(\partial_z^3\psi_t)(U_t)/\psi_t'(U_t)$. Since $dU_t=\sqrt{8/3}\,dB_t$, the supplied identity and <Itô formula> give
$$
d\log J_t
=q_t\,dU_t
+\left[-\frac56q_t^2
+\left(-\frac43+\frac12\frac83\right)r_t\right]dt
=q_t\,dU_t-\frac56q_t^2dt.
$$
For $M_t=J_t^\alpha$, its drift coefficient is
$$
-\frac{5\alpha}{6}+\frac12\alpha^2\frac83
=\frac{\alpha(8\alpha-5)}6.
$$
Thus the nonzero choice is
$$
\boxed{\alpha=\frac58,}
$$
and $M_{t\wedge\tau}$ is a continuous local martingale. The boundary Schwarz lemma for mapping-out maps gives $0\leq J_{t\wedge\tau}\leq1$, so $0\leq M_{t\wedge\tau}\leq1$. A bounded local martingale is a true martingale. This is the <SLE eight-thirds restriction martingale>.
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