Suppose I have the following function where $$z=\omega(\zeta)=\frac{1}{\zeta}$$ and also, $$\phi(\zeta) = \zeta^{-1}+2\zeta$$ By using chain rule, I can get the first-derivative of $\phi(z)$. Notice that I want now $\phi$ as a function of $z$. Thus, $$\phi'(z)=\frac{d\phi}{d\zeta}\frac{d\zeta}{dz}=\frac{\phi'(\zeta)}{\omega'(\zeta)}$$ where, $$\omega'(\zeta) = \frac{d\omega(\zeta)}{d\zeta}=\frac{dz}{d\zeta}$$
My question is, how can we get the second-derivative of $\phi(z)$, i.e. $\phi''(z)$? By using chain rule?
You already have $\phi'(z)$, so just differentiate it using the product and chain rules:
$$\phi''(z) = \frac{d}{dz}\left(\frac{d\phi}{d\zeta}\right)\frac{d\zeta}{dz} + \frac{d\phi}{d\zeta}\frac{d}{dz}\left(\frac{d\zeta}{dz}\right) = \frac{d^2\phi}{d\zeta^2}\left(\frac{d\zeta}{dz}\right)^2+\frac{d\phi}{d\zeta}\frac{d^2\zeta}{dz^2}.$$
Notice that nothing we've done uses the form of your functions, so this holds in full generality. In your special case, you could also just substitute everything into get $\phi$ in terms of $z$ directly, then differentiate that twice.