Vrftjkry

I have the following O-U process:
$$d log z_t =-nu log z_t d_t + barsigma dW_t tag{1}$$

and want to apply Itô's formula as:

$$dy=df(x)=bigg(mu(x)f'(x)+frac{1}{2}sigma^2(x)f''(x) bigg)dt + sigma (x)f'(x) tag{2}$$

Since the process is logged, I tried considering a non logged version as

$$d z_t =-nu z_t d_t + sigma dW_t tag{3}$$

and applying the transformation $Z_t=log z_t$. Thenby applying Itô's formula I obtain:

$$ begin{aligned} d Z_t &= bigg(-nu z_t frac{1}{z_t} + frac{1}{2}sigma^2 frac{-1}{z_t^2}bigg)dt+sigma frac{1}{z_t}dW\
&=bigg(-nu-frac{sigma^2}{2cdot z_t^2} bigg)dt+frac{sigma}{z_t} d Wend{aligned} tag{4}$$

Going back to logs:

$$ log z_t=bigg(-nu-frac{sigma^2}{2cdot z_t^2} bigg)dt+frac{sigma}{z_t} d Wtag{5}$$

However, I was supposed to obtain

$$mu(z)=bigg(-nu log(z)+frac{bar{sigma}^2}{2} bigg)z, text{ }sigma(z)=barsigma z tag{6}$$

which I do not see.

Any help is highly appreciated

Edit: The solution is given Did's hint:

Define a flexible guess for $d z_t$ as
$$dz_t=mu(z_t)dt+sigma(z_t)dW$$
I now define $y(t)=log(z_t)$ and apply Itô's formula to the above as
$$dlog z_t = bigg(mu(z_t)frac{1}{z_t}-frac{1}{2}sigma^2(z_t)frac{1}{z^2} bigg)dt + sigma (z_t)frac{1}{z_t}$$
Recalling that I have the definition of $d log z_t$ from (1), for the drift coefficient
$$begin{aligned}-nu z_tlog z_t dt&= bigg(mu(z_t)frac{1}{z_t}-frac{1}{2}sigma^2(z_t)frac{1}{z_t^2} bigg)dt\
Leftrightarrow mu(z_t)&=-nu z_tlog(z_t)+frac{sigma^2}{2 z_t}end{aligned}tag{*}$$
Solving for the diffusion coefficient yields
$$barsigma=sigma(z_t) frac{1}{z_t} Leftrightarrow sigma(z_t)=bar{sigma}z_t $$
such that (*) becomes
$$begin{aligned}mu(z_t)&=-nu z_t log z_t+frac{(bar{sigma}z_t)^2}{2 z_t}\
&=bigg(-nu log z_t +frac{bar{sigma}^2}{2}bigg)z_t
end{aligned}$$