Vrftjkry

Suppose I have the following function where
$$z=omega(zeta)=frac{1}{zeta}$$ and also,
$$phi(zeta) = zeta^{-1}+2zeta$$
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,
$$phi'(z)=frac{dphi}{dzeta}frac{dzeta}{dz}=frac{phi'(zeta)}{omega'(zeta)}$$
where,
$$omega'(zeta) = frac{domega(zeta)}{dzeta}=frac{dz}{dzeta}$$

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:

$$phi''(z) = frac{d}{dz}left(frac{dphi}{dzeta}right)frac{dzeta}{dz} + frac{dphi}{dzeta}frac{d}{dz}left(frac{dzeta}{dz}right) = frac{d^2phi}{dzeta^2}left(frac{dzeta}{dz}right)^2+frac{dphi}{dzeta}frac{d^2zeta}{dz^2}.$$

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.