I have a problem
$k[[x,y]]/I$ is a Gorenstein ring implies that $I$ is generated by 2 elements.
I am stuck since I do not have many techniques to prove that an ideal is generated by 2 elements.
Here is what I have, denote $R=k[[x,y]]$, since $R$ is a regular local ring, I can apply Auslander-Buchsbaum formula: $\operatorname{pd}_R R/I=\operatorname{depth} R-\dim_R(R/I)=2$. I thought Hilbert-Burch theorem could help but still, I'm stuck.
Thank you for your help
Standard arguments go as follows. You did not explicitly say that $I$ has height two, though you use it in your Auslander-Buchsbaum.
In this case, Gorenstein is equivalent to saying $\mathrm{Ext}^1(I,R)=R/I$ (Do you know why?) Then $1\in R/I$ gives you an extension $0\to R\to P\to I\to 0$. Then depth of $P$ is at least one, since both $R,I$ are. So, projective dimension of $P$ is at most one. But, $\mathrm{Ext}^1(P,R)=0$, by our choice of the extension and thus we see that projective dimension of $P=0$, which means it is free and thus free of rank 2. Now it is immediate that $I$ is two generated. This proof is usually called `Serre's construction'.