Special change of coordinates for 1-form Consider the curve $C=Z(-x_1^5+x_0^{4}x_2-x_2^5) \subset \mathbb{P}^2$. Dehomogenize with respect to $x_2$ to obtain an irreducible affine curve $U_2=Z(-y^5+x^4-1)$. Notice that $C=U_2 \cup P$ where $P$ is the point at infinity $(1:0:0)$. Define a 1-form on $U_2$:
\begin{equation}
\omega= \frac{dx}{5y^4}=\frac{dy}{4x^3}.
\end{equation}
I want to study this differential form at the point $P$. With a change of coordinate $x=v^{-1}$, $y=uv^{-1}$, I need to express $\omega$ at $P$ in terms of $du$ $\textit{only}$ (multiplied by an appropriate non-vanishing rational function, as $u$ is a uniformising parameter in this context, if you are familiar with the notion) but when I try to calculate $du$ and $dv$ from $dx$ and $dy$, I inevitably obtain an expression that contains both $du$ and $dv$ and I don't know how to get rid of the $dv$. I would like to know whether I am miscalculating or there is another way to obtain that expression for $\omega$. 
Edit: if I try to write the first form $\omega$ in terms of $u,v$ I get:
\begin{equation}
\frac{-(v^2dv)}{5u^{4}} = \frac{-5u^4v^{2} du}{5u^4(1-5v^4)} = \frac{-v^{2} du}{1-5v^4}
\end{equation}
but this rational function vanishes at $P$ which should not happen.
My end goal is to write $\omega = f u^n \, du$ with $n \in \mathbb{Z}$ and $f$ a rational function with $f(P) \neq 0$. 
 A: Here's a summary of the comments.
Rewriting the equations for $U_2$ in terms of $u$ and $v$, we find
$$
\left(\frac{u}{v}\right)^5 = \frac{1}{v^4} - 1 \implies u^5 = v - v^5
$$
and thus obtain an affine curve $U: u^5 = v - v^5$. Since $x = X_0/X_2$ and $y = X_1/X_2$, then
$$
v = \frac{1}{x} = \frac{X_2}{X_0} \quad \text{and} \quad u = yv = \frac{y}{x} = \frac{X_1}{X_0}
$$
so $P = (1:0:0)$ corresponds to $(u,v) = (0,0)$. In $(u,v)$ coordinates, $\omega$ transforms into
$$
\omega = \frac{dx}{5y^4} = \frac{d(1/v)}{5(u/v)^4} = \frac{-\frac{1}{v^2} dv}{5 \frac{u^4}{v^4}} = -\frac{v^2}{5 u^4} dv
$$
Differentiating the equation for $U$, we find $5u^4 du = dv - 5v^4 dv = (1-5v^4)dv$, so
$$
dv = \frac{5u^4}{1 - 5 v^4} du \, .
$$
Thus
$$
\omega = -\frac{v^2}{5 u^4} \frac{5u^4}{1 - 5 v^4} du = -\frac{v^2}{1 - 5 v^4} du \, .
$$
Let $\DeclareMathOperator{\m}{\mathfrak{m}} \m = (u,v)$ be the maximal ideal of the local ring at $P$ (so $\m = (u)$ since $u$ is a uniformizer at $P$), and let $\newcommand{\ord}{\operatorname{ord}_P} \ord$ be the associated discrete valuation (the order of vanishing). Since $v = u^5 + v^5$, we see that $v \in \m^5$, so $\ord(v) \geq 5$. Then $\ord(v^5) \geq 25$ and since $\ord(u^5) = 5$, then $\ord(u^5) \neq \ord(v^5)$, and hence
$$
\ord(v) = \min\{\ord(u^5), \ord(v^5)\} = 5
$$
by properties of discrete valuations (see here). Thus $v^2$ has valuation $10$, so $v^2$ and $u^{10}$ differ multiplicatively by a unit. To find this unit, we square the equation for $U$:
\begin{align*}
u^{10} = (v - v^5)^2 = v^2 - 2v^6 + v^{10} = v^2(1 - 2v^4 + v^8)
\end{align*}
and find that $v^2 = \frac{u^{10}}{1 - 2v^4 + v^8}$. Thus
$$
\omega = -\frac{v^2}{1 - 5 v^4} du = -\frac{\frac{u^{10}}{1 - 2v^4 + v^8}}{1 - 5 v^4} du = -\frac{1}{(1 - 2v^4 + v^8)(1 - 5 v^4)} u^{10} \, du
$$
as desired.
