Is the function ring $C^{\infty}(M)$ noetherian? Let $M$ be a smooth manifold and $C^{\infty}(M)$ be its function ring. Is this a noetherian ring?

No, the ring $C^{\infty}(M)$ is never noetherian if $\dim M\gt 0$.
Indeed consider a strictly decreasing sequence of closed subsets $$M=C_0\supsetneq C_1 \supsetneq C_2 \supsetneq C_3\supsetneq  \cdots$$ (Such a sequence is easy to construct: take the inverse image in a  chart of closed concentric balls  in $\mathbb R^n$.)
A theorem of Whitney ensures that there exist a smooth function $f_j\in C^{\infty}(M)$ whose zero locus is precisely $C_j$.
Now if you consider the ideal $I_j=Z(C_j)\subset C^{\infty}(M)$ of functions zero  on $C_j$ you get a strictly increasing sequence of ideals of $C^{\infty}(M)$ destroying all hopes of noetherianness for that ring: $$I_0 \subsetneq I_1 \subsetneq I_2 \subsetneq I_3 \subsetneq  \cdots \subsetneq C^{\infty}(M)\quad (f_{j+1}\in I_{j+1}\setminus I_j)$$
