Let $R=\mathbb C[x,y]/(y(x-a)(x-b)-1)$ where $a,b$ are distinct complex numbers. Show that the cohomology of the de Rham complex $$0\to R\to \Omega_{R/\mathbb C}\to 0$$ is $\mathbb C$ in degree zero and $\mathbb C^2$ in degree one.
I have no experience in computing de Rham cohomology whatsoever (maybe except the case $R=\mathbb C[x]$, not with quotients), but first I need to at least identify $\Omega _{R/\mathbb C}$ and to write down to write the map $R\to \Omega_{R/\mathbb C}$. How do I do that?
You have $R=\Bbb C[x,y]/(F(x,y))$ for your given polynomial $F(x,y)$. The differentials $\Omega_{R/\Bbb C}$ form an $R$-module, generated by symbols $dx$ and $dy$ subject to the single relation $F_x(x,y)\,dx+F_y(x,y)\,dy=0$ where $F_x$ and $F_y$ are partial derivatives. The map $d:R\to\Omega_{R/\Bbb C}$ takes $g(x,y)$ to $g_x(x,y)\,dx+g_y(x,y)\,dy$.
The kernel of $d$ certainly contains the elements of $\Bbb C$. What you now need to do next is to show that if $g$ is a polynomial with $(g_x,g_y)=(hF_x,hF_y)$ for some polynomial $h$, then $g\in\Bbb C$. Over to you!