How to compute this $\mathrm{Ext}^1$? Let $A$ be a regular local $\mathbb{C}$-algebra of dimension $2$, such as the localization of $\mathbb{C}[x,y]$ at $(x,y)$, and let $\nu=(\nu_1\geq\nu_2\geq\cdots\geq\nu_{\ell}\geq0)$, $\mu=(\mu_1\geq\mu_2\geq\cdots\geq\mu_m\geq0)$ be partitions of the same $n\in\mathbb{N}$.  Consider the ideals $I=(y^{\nu_1},xy^{\nu_2},x^2y^{\nu_3},\ldots,x^{\ell - 1}y^{\nu_\ell},x^\ell)$ and $J=(y^{\mu_1},xy^{\mu_2},x^2y^{\mu_3},\ldots,x^{m-1}y^{\mu_m},x^m)$ generated by elements in the maximal ideal (here $x,y$ are the uniformizers). So, $A/I$ and $A/J$ have the same length equal to $n$.

How to compute $\mathrm{Ext}^1_A(A/I,A/J)$ in terms of $\nu$ and $\mu$?

Edit: I would add that it's probably possible to make the computation using the theory of ADHM data developped e.g. in Nakajima's Lectures on Hilbert schemes of points on surfaces. But I would like to get a direct answer that does not use ADHM data and such.
 A: First off, Ext and localisation commute in situations like this (Weibel An introduction to homological algebra Proposition 3.3.10 p.76) so we can work with $A=k[x,y]$ if we like.  I will drop the assumption that the two partitions have the same size, not least because you get a nice interpretation of $\operatorname{Ext}^1$ when the first partition is (1): it has a basis indexed by addable nodes of the second partition.
Let 
$\lambda=\lambda_1^{a_1} \lambda_2^{a_2}\cdots$
be a partition of $n$, where
$\lambda_1>\lambda_2>\cdots> \lambda_l >0$.
A minimal $A$-projective resolution of the module you associate to $\lambda$ is
$$0 \to \bigoplus_{i=1}^l A \otimes (x^{a_i}\otimes x^{\sum_{j=0}^{i-1}a_j } y^{\lambda_i}-y^{\lambda_i-\lambda_{i+1}}  \otimes x^{\sum_{j=0}^i a_j}y^{\lambda_{i+1}})
\to \bigoplus _{i=0}^l A\otimes x^{\sum_{j=0}^i a_j}y^{\lambda_{i+1}}
\to A \to A/I_\lambda \to 0$$
in which you should interpret $\lambda_{l+1}$ as zero.
This leads to a direct but awkward description of $\operatorname{Ext}^1_A(A/I_\lambda, A/I_\mu)$ as follows.  Consider the Ferrers diagram of $\mu$ drawn in the French style: by this I mean the subset of $\mathbb{Z}\times \mathbb{Z}$ consisting of $(0,0), (0,1), \ldots, (0,m_1)$, $(1,0), (1,1), \ldots, (1,m_2)$ and so on, where $\mu=(m_1,m_2,\ldots)$.
Pick $0 \leq r \leq s \leq l$ and choose elements $\rho_s,\ldots,\rho_s$ of the Ferrers diagram of $\mu$ with the property that 
$\rho_r + (0,\lambda_r-\lambda_{r-1}) \notin \mu$, 
$\rho_j + (a_{j+1},0)=\rho_{j+1} + (0,\lambda_{j+1}-\lambda_j) \in \mu$ for each $r<j<s$, and
$\rho_s + (a_{s+1},0) \notin \mu$.
The first condition should be dropped if $r=0$ and the last if $s=l$.
Thus these elements form the concave corners in a fragment of the outer rim of the Ferrers diagram for $\lambda$.  The Ext-group is the free vector space on these "paths," modulo linear combintations of paths that arise from the entire outer rim of the Ferrers diagram for $\lambda$.
Things look a little nicer in the case $\lambda = (1)$.  Then a path with $r=s=0$ is an element of the Ferrers diagram of $\mu$ with nothing directly to its right, a path wtih $r=s=1$ is an element with nothing directly above it, and every path with $r=0,s=1$ is a coboundary.  Furthermore an $r=s=1$ path is a boundary if there's an element of the diagram to its left and a $r=s=0$ path is a boundary if there's an element of the diagram above it.  It follows the non-bounding paths sit next to the addable nodes of $\mu$.
