I think there is no such $f$ in general unless there's some specifics to your problem that I'm missing.

We need $\sigma^2 f'(\mu)^2$ to not depend on $\sigma^2$. To try to formalize this, I'll interpret "not depend" to mean 
$$
\frac{\partial}{\partial \sigma^2} [\sigma^2 f'(\mu)^2] = 0.
$$

In a typical Gaussian distribution $\mu$ and $\sigma^2$ are free to vary separately and neither is viewed as a function of the other, i.e. $\frac{\partial \mu}{\partial \sigma^2} = 0$. Furthermore you write $f$ and $f'$ as only functions of $\mu$, and in general I think it'd be very artificial to have them depend on $\sigma$ since then $f$ isn't a statistic, so I'll also assume that $f$ and $f'$ do not have $\sigma$ in them.

This means the derivative is just
$$
f'(\mu)^2 = 0 \implies f'(\mu) = 0 \text{ everywhere}.
$$
But functions with $f'(\mu) = 0$ are excluded from the delta method by hypothesis so there are no valid functions that satisfy our requirements here.