Quotientring-Show the isomorphism To show that $\mathbb{C}[x,y]/\langle x,y \rangle \cong \mathbb{C}$ I did: 
$p(x,y) \in \mathbb{C}[x,y]$ 
$p(x,y)=\sum_{m,n=0}^k a_{mn}x^m y^n$ 
modulo $\langle x, y \rangle$ we have that $x \equiv 0$ and $y \equiv 0$ 
So $p(x,y)=a_{00} \pmod {\langle x, y \rangle } \in \mathbb{C}$ 
Is my approach correct? What can I improve?

What you've done is pretty much the first isomorphism theorem: given a ring homomorphism $\phi : R \to S$, we have the isomorphism
$$
\frac{R}{\ker(\phi)} \cong \text{im}(\phi).
$$
So to write out what you thought a bit more explicitly, take $\phi : \Bbb C[x,y] \to \Bbb C$, with
$$
\phi\left(\sum_{m,n=0}^k a_{mn}x^my^n\right) = a_{00}.
$$
Then the kernel of $\phi$ consists of all the polynomials with no constant term, that is, $\langle x,y \rangle$. The image of $\phi$ is clearly the whole of $\Bbb C$, so the result follows from the isomorphism theorem.
