Let $A$ be $C^*$-algebra, $ x\in A$ be normal, and $f\in C(sp(x))$. Show that $(g\circ f)(x)=g(f(x))$ for all $g\in C(f(sp(x)))$.
My attempt: I think it suffices to show that $C(f(sp(x))) \subseteq C(sp(x))$.
Consider $$C(f(sp(x))) =C(sp(f(x)))=C^*(f(x),1) \subseteq C^*(x,1)=C(sp(x)).$$
Is the above correct? If yes, can someone help me justify the second inequality and the inclusion? If not, then how can I prove the given statement?
The inclusion you give is not true: consider for instance $\operatorname{sp}(x)=[0,1]$, and $f(t)=t+2$. Then $f(\operatorname{sp}(x))=[2,3]$, and you cannot say that $C[2,3]\subset C[0,1]$ (though they are isomorphic in this case; but in general, they will not be).
In any case, it looks like you are missing the point. The left-hand-side is the functional calculus of $x$ through the function $g\circ f$. The right-hand-side is the functional calculus of $f(x)$ (which is the functional calculus of $x$ through $f$) through the function $g$. It is not obvious that those two things are the same, though they are, and that's what you need to prove.