I need to evaluate left derived functors of $\mathrm{Ext}^2_{\mathbb{C}[x,y]}(\mathbb{C}[x,y]/(x^2,xy,y^2), -)$ on $\mathbb{C}[x,y]/(x,y)$ but i have no idea how to evaluate zeroth functor..
I wrote kozul's resolvent $P^\bullet$ for $\mathbb{C}[x,y]/(x,y))$ so $\mathrm{Ext}^2_{\mathbb{C}[x,y]}(\mathbb{C}[x,y]/(x^2,xy,y^2), \mathbb{C}[x,y]/(x,y))$ is zeroth cohomology of $\mathrm{Ext}^2_{\mathbb{C}[x,y]}(\mathbb{C}[x,y]/(x^2,xy,y^2), P^\bullet)$.
But also i can't evaluate $\mathrm{Ext}^2_{\mathbb{C}[x,y]}(\mathbb{C}[x,y]/(x^2,xy,y^2), \mathbb{C}[x,y])$.
Could you give me some hints?
Denote $R = \mathbf{C}[x,y]$. First we find a free resolution for $R/(x^2,xy,y^2)$:
$$0 \longrightarrow R^{\oplus 2} \xrightarrow{\begin{bmatrix} y & 0\\ -x & y\\ 0 & -x \end{bmatrix}} R^{\oplus 3} \xrightarrow{\begin{bmatrix} x^2 & xy & y^2 \end{bmatrix}} R \longrightarrow 0.$$ Note that this is exact since the relations on $m_1 = x^2$, $m_2 = xy$, and $m_3 = y^2$ are generated by $f_1 = ym_1 - xm_2$, $f_2 = ym_2 - xm_3$, and $f_3 = y^2m_1 - x^2m_3$, but this last relation is generated by the first two: $f_3 = yf_1 + xf_2$.
Now applying $\operatorname{Hom}\!\left(-,R/(x,y)\right)$ gives the chain complex $$0 \longleftarrow \left(R/(x,y)\right)^{\oplus 2} \overset{0}{\longleftarrow} \left(R/(x,y)\right)^{\oplus 3} \overset{0}{\longleftarrow} \left(R/(x,y)\right) \longleftarrow 0$$ since all the matrices are zero mod $(x,y)$. Thus, $$\operatorname{Ext}^2_R\!\left(R/(x^2,xy,y^2),R/(x,y)\right) = \left(R/(x,y)\right)^{\oplus 2}.$$