= Solution
Write the <semimartingale decomposition> as $U=M+A$, where $M$ is a continuous local martingale and $A$ has finite variation. Applying <Itô formula> to $\log Z_t(z)$ shows that its finite-variation part is
$$
\frac{2}{Z_t(z)^2}dt
-\frac1{Z_t(z)}dA_t
-\frac1{2Z_t(z)^2}d\langle M\rangle_t.
$$
It vanishes for every $z$. Multiplying by $Z_t(z)^2$ gives
$$
2dt-Z_t(z)dA_t-\frac12d\langle M\rangle_t=0.
$$
Subtract this identity for two points with distinct $Z_t$ to obtain $dA_t=0$; then $d\langle M\rangle_t=4dt$. Since the curve starts at zero, $U_0=0$. The <Lévy characterization of Brownian motion> now gives $U_t=2B_t$. Hence the Loewner chain is
$$
\boxed{\operatorname{SLE}_4.}
$$
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