Does there exist a stochastic time derivative?

I think the chain rule $d[f(Z_t)]=f'(Z_t)\circ dZ_t$ is valid when the product $\circ$ is defined as in Stratonovich stochastic integral (while the SDE uses Ito's). Note that $g(Z_t)\circ dt$ doesn't differ from the ordinary product $g(Z_t)\ dt$, but $g(W_t)\circ dW_t$ is Ito's $g(W_t)\ dW_t$ plus $\frac12 g'(W_t)\ dt$.

The Malliavin derivative satisfies the requisite chain rule. However, if $Z_t$ is no longer a function of the Wiener process, then the Malliavin derivative of $f(Z_t)$ is identically zero.

To see this, we briefly recall the nice link between the Girsanov theorem and Malliavin calculus. Let $\epsilon>0$ be a perturbation parameter and let $\phi \in L^2([0,T]; \mathbb{R}^d)$. Recall that Girsanov theorem says that $Z_t$ (given by the OP) on the probability space $(\Omega, \mathcal{F}, P)$ is a weak solution to $$ dZ_t = \mu(t,Z_t) dt + \sigma(t,Z_t) \left( d W_t + \epsilon \phi(t) dt \right) $$ on the probability space $(\Omega, \mathcal{F}, \tilde P)$ where $$ \frac{d \tilde P}{dP}= \exp\left( -\frac{\epsilon^2}{2} \int_0^T | \phi(s)|^2 ds + \epsilon \int_0^T \phi(s) dW(s) \right) $$ More to the point, the variations taken in the Malliavin derivative are precisely of this form: the Wiener process is varied arbitrarily in the direction of the deterministic $\phi(\cdot)$. From this link we immediately see that if $\sigma=0$, then these variations have no effect on $f(Z_t)$.

Reference

Nualart, D. (2006). The Malliavin calculus and related topics. Springer.