We are working over an algebraically closed field $k$ of characteristic $0$. Let $X^o = V(x^6 + xy^5 + y^6 − 1) \subset \mathbb{A}^2_k$ and let $X \subset \mathbb{P}^2_k$ be the projective closure. I have shown that $X$ is a smooth curve by computing the dimension of tangent spaces at each point.
The next thing the problem wants me to do is show that $\omega = dx/(5xy^4 + 6y^5)$ is a regular differential on $X$. Away from the infinity, I took $x-x(p)$ and $y-y(p)$ as local parameters (depending on whether $\partial f / \partial x$ and $\partial f / \partial y$ are $0$ or not, $f$ is the polynomial from above) and then concluded that the order $\nu_p(\omega) = 0$ here. I hope someone can confirm that this is still fine.
Now I am less sure about the continuation. If we were using the coordinates $[x : y: 1]$ earlier, let's now use $[u : 1 : v]$. They have to be compatible on the intersection, so we get the relationship $u=x/y$ and $v=1/y$. This allows us express $\omega$ in terms of the new coordinates after making the substitution and using the chain rule. In theory, we have now translated the problem to a new local affine chart and could do the same as above. The problem is that I can somewhere use $v$ as a local parameter and then get $\nu_p(\omega)=\nu_p(-(5u^4+6u^5)/v^3)=-3$, but the differential is supposed to be regular, so I am doing something wrong.