It's well know from Functional Analysis that if $T$ is a operator in a $\mathbb{C}$-vector space and $p\in\mathbb{C}[z]$, then $\sigma(p(T))=p(\sigma(T))$.
It's true that for $p,q\in\mathbb{C}[z]$, $f=\frac{p}{q}$, where $q$ has not root in $\sigma(T)$ yields $\sigma(f(T))=f(\sigma(T))$?
I thought, how $\sigma(p(T))=p(\sigma(T))$ and $\sigma(q(T))=q(\sigma(T))$, then taking $\lambda\in\sigma(p(T))$ and $\mu\in\sigma(p(T))$ so $\frac{\lambda}{\mu}\in\sigma(f(T))$ is by this way to proof?
If $q$ does not vanish on $\sigma(T)$ then $q(T)$ is invertible. Then $$f(T)-\lambda I=p(T)q(T)^{-1}-\lambda I=q(T)^{-1}[p(T)-\lambda q(T)]$$ The operator $p(T)-\lambda q(T)$ is noninvertible iff $$0\in \sigma(p(T)-\lambda q(T))=(p-\lambda q)(\sigma(T))$$ Equivalently $p(z)-\lambda q(z)$ vanishes on $\sigma(T),$ i.e. $p(z_\lambda)-\lambda q(z_\lambda)=0$ for some $z_\lambda\in\sigma(T).$ Therefore $$\lambda={p(z_\lambda)\over q(z_\lambda)}=f(z_\lambda)$$ hence $\lambda\in f(\sigma(T)).$ The proof shows $$\sigma(f(T))= f(\sigma(T))$$