Compute a 2-differential form Let $J$ be the open interval $J = (0, 1) \subset \mathbb{R}$
Let the 1-differential form $\alpha =\Big( \dfrac{\log(x)}{1-y} + \dfrac{\log(1-x)}{y} \Big) \in \Omega^0(J^2)$
I want to compute $d\alpha$
I suppose I'd have to use the formula $d\alpha = \dfrac{\partial\alpha}{\partial x}dx + \dfrac{\partial\alpha}{\partial y}dy $
Heres's my attempt
$\dfrac{\partial\alpha}{\partial x}dx + \dfrac{\partial\alpha}{\partial y}dy = \Big(\dfrac{1}{x-xy} - \dfrac{1}{y-xy}\Big)dx + \Big(\log(x)(-\dfrac{1}{(1-y)^2}\log(x) - \dfrac{1}{y^2}\log(1-x)\Big)dy$ 
Of course, I feel like something is wrong.
Edit 
Now I've almost entirely changed my post, but how would I compute $d\alpha$ if alpha was indeed a 1-form and not a 0-form. ($\alpha =\Big( \dfrac{\log(x)}{1-y} + \dfrac{\log(1-x)}{y} \Big)dy)$
 A: It is nearly correct. You have the $0$-form (i.e. a function) $\alpha =\Big( \dfrac{\log(x)}{1-y} + \dfrac{\log(1-x)}{y} \Big)$ and you want to compute the $1$-form $\text{d}\alpha$. Be clear that the exterior differential (one also calls this Cartan differential) of a $p$-form is given by a $(p+1)$-form. This is immediate from the representation theorem for $p$-forms.
Yes, $\text{d}\alpha = \dfrac{\partial\alpha}{\partial x}dx + \dfrac{\partial\alpha}{\partial y}dy$ is correct.
About your attempt
$$\dfrac{\partial\alpha}{\partial x}\text{d}x + \color{red}{\dfrac{\partial\alpha}{\partial y}}\text{d}y = \Big(\dfrac{1}{x-xy} - \dfrac{1}{y-xy}\Big)\text{d}x + \Big(\color{red}{\log(x)(-\dfrac{1}{(1-y)^2}\log(x) - \dfrac{1}{y^2}\log(1-x)}\Big)\text{d}y$$
The red thing is not correct since $\left(\frac{1}{1-y} \right)'=\frac{1}{(1-y)^2}$ because of the chain rule and therefore you get 
$$\partial_y \alpha=\frac{\log(x)}{(1-y)^2}-\frac{\log(1-x)}{y^2}$$
instead of the red part above. Additionally you messed up something in the red part with the parenthesis.
EDIT: To answer the second part, if you have a $1$-form $$\beta =\Big( \dfrac{\log(x)}{1-y} + \dfrac{\log(1-x)}{y} \Big)dy=:B\ \text{d}y$$ then its exterior differential is given by the $2$-form 
$$\text{d}\beta=\text{d}B \wedge \text{d}y=\partial_x B \ \text{d}x \wedge \text{d}y +  \partial_y B \ \text{d}y \wedge \text{d}y=\partial_x B \ \text{d}x \wedge \text{d}y$$ 
where we used that $\text{d}{B}=\partial_x B \ \text{d}x+\partial_y B \ \text{d}y$ (as seen above for $\alpha$) and $\text{d}y \wedge \text{d}y=0$.
