= Solution
Consider the subalgebra
$$
\boxed{A=\mathbb C+x\mathbb C[x,y]=\mathbb C[x,xy,xy^2,xy^3,\ldots]\subseteq\mathbb C[x,y].}
$$
It is closed under multiplication because every nonconstant term of its elements is divisible by $x$. Evaluation at $x=0$ defines a homomorphism $A\to\mathbb C$ with kernel $I=x\mathbb C[x,y]$. Its square is $I^2=x^2\mathbb C[x,y]$, so
$$
I/I^2\cong x\mathbb C[y]
$$
as a complex <vector space>. In particular, the classes of $xy^j$, $j\ge0$, are linearly independent and span it.
If $A$ were a <finitely generated algebra>, take a finite list of generators and subtract their scalar images under evaluation. The resulting generators $z_1,\ldots,z_r$ lie in $I$ and still generate $A$ over $\mathbb C$. Every element of $I$ is a polynomial in the $z_i$ with zero constant term. Modulo $I^2$, all terms of degree at least two vanish, leaving a linear combination of the classes of $z_1,\ldots,z_r$. This would make $I/I^2$ finite-dimensional, contradicting the displayed infinite basis. Thus \b[this subalgebra is not finitely generated]. This is the <infinite cotangent space obstructs finite algebra generation> argument; it proves failure of finite generation even when the proposed generators are arbitrary polynomials, rather than just monomials.
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