I'm studying algebraic geometry and we've just been introduced to differential forms. The book (Algebraic Geometry by Garrity .et al) gives us this exercise:
Exercise 3.6.33: Suppose $x$ and $y$ are both local coordinates at a point $p$ on a smooth curve $C$. We will show that the divisor of a differential form $\omega$ does not depend on the choice of local coordinates.
(1) Suppose $x$ and $y$ are coordinates in the same affine patch of $\mathbb{P}^2$. We can describe $C$ near $p$ by $P(x, y) = 0$. Use this to prove the divisors $dx$ and $dy$ both have order zero at $p$. [Hint: Since $P(x, y) = 0$, we have $0 = dP = P_x dx + P_y dy$.]
(2) Suppose $x$ and $y$ are coordinates in two different affine patches. We can describe $C$ near $p$ by either $P(x, u) = 0$ or $P(y, v) = 0$, where $(x, u)$ and $(y, v)$ are two affine charts of $\mathbb{P}^2$. Use that $x$ and $u$ can each be written as rational functions in $y$ and $v$ to show that the divisors of $dx$ and $dy$ both have order zero at $p$. [Hint: This is an exercise in the chain rule. Let $f(y, v)$ be the rational function with $x = f(y, v)$. Then $1 = \frac{dx}{dx} = \frac{dy}{dx} = f_y y_x + f_v v_x$. Also, $dx = f_y d_y + f_v dv$, which can be written as a rational function times $dy$, since $dv = -\frac{P_y}{P_v} dy$.
You will also need to show and use $-\frac{P_y}{P_v} = \frac{v_x}{y_x}$.]
Here is my attempt to solve them:
Since $x$ and $y$ are local coordinates, so $P_y \neq 0$ and $P_x \neq 0$ and so $\frac{P_x}{P_y} \neq 0$.
$$ dx = f_y dy + f_v dv $$ $$ = \left(f_y - f_v \frac{P_y}{P_v}\right)dy $$ $$ = \left(\frac{df}{dy} - \frac{df}{dv}\frac{dP/dy}{dP/dv}\right)dy $$ $$ = \left(\frac{df}{dy} - \frac{df}{dv}\frac{dv}{dy}\right)dy $$ $$ = 0 dy $$ and so $div(0) = 0$. Is this answer correct, or what is wrong with my reasoning?