Let $R = \mathbb{C}[x,y]$. Then $R/Rx$ is isomorphic to $\mathbb{C}[y]$.
My Proof
$R = ax+by+c , a,b,c \in \mathbb{C}$ and $Rx = (ax+by+c)x , a,b,c \in \mathbb{C}$
So, $$R/(R\cdot x) = (ax+by+c) + (a'x+b'y+c')x, a,b,c a',b',c' \in \mathbb{C}$$ $$ = by+c + (a'x+b'y+(c'+a))x, a,b,c a',b',c' \in \mathbb{C}$$ So the set of cosets of $R/(R\cdot x)$ is clearly isomorphic to $\mathbb{C}[y]$.
This argument feels incredible handwavy. Is there a more rigorous way to prove this?
(I know in some cases defining an appropriate ring homomorphism and finding the kernel gives us the isomorphism by the first isomorphism theorem, but I couldn't find a homomorphism that worked.)
Hint. Can you prove that \begin{align*} \Phi:\mathbf{C}[X,Y] &\longrightarrow \mathbf{C}[Y] \\ f(X,Y) &\longmapsto f(0,Y) \end{align*} is a surjective ring homomorphism? Its kernel is exactly $(X)\subset \mathbf{C}[X,Y]$ (why?). Now you are ready to use the first isomorphism theorem.