 Solution

Yes. Taking the constant coefficient defines a surjective ring homomorphismIts kernel of a ring homomorphism is $(x)$, so the first isomorphism theorem for rings gives $R[[x]]/(x)\cong R$.

We prove the needed preservation result. If $A$ is a Noetherian ring and $J\subseteq A$ is an ideal, any ascending chain of ideals in $A/J$ lifts under the quotient ring homomorphism to an ascending chain of ideals in $A$. The lifted chain stabilizes; surjectivity makes the original chain stabilize as well. Hence every quotient ring of a Noetherian ring is Noetherian.

Applying this with $A=R[[x]]$ and $J=(x)$ gives