My question in short : Let $X$ be a smooth projective hypersurface, $\omega$ be a meromorphic form on $\mathbb{CP}^n$ with a simple pole along $X$. How to prove the independence of coordinate of the Poincaré Residue map using local coordinates?
More details : Given a hypersurface $X\subset \mathbb{CP^n}$, defined by a degree $d$ polynomial $F$, a meromorphic form $\omega$ with pole of order 1 along $X$. Consider a chart on $\mathbb{CP}^n$ containing $X$ as the vanishing locus of $f$, we can write $\omega=\dfrac{df}{f}\wedge \rho$ and define : $$\text{Res}(\omega)=\text{Res}(\dfrac{df}{f}\wedge \rho)=\rho|_{X}.$$ I would like to prove that this definition of $\text{Res}(\omega)$ is independent of chart, (thus really a global form on $X$.)
Background : The construction above is in the wiki page of Poincaré Residue. This construction is also mentioned in page 147 of Griffiths-Harris. I was originally reading the book Enumerated Geometry and String theory by Katz. In page 175, the author attempts to show that a smooth quintic threefold is Calabi-Yau, using elementary local coordinates (without using the adjunction formula).
In this case, $d=5$, $n=4$ and $\omega=\dfrac{\Lambda}{F}$, where $$\Lambda=x_0dx_1\wedge dx_2\wedge dx_3\wedge dx_4 + \dots + x_4dx_0\wedge dx_1\wedge dx_2\wedge dx_3\wedge dx_4,$$ obtained by cyclically permuting the indices $0,\dots,4$, and $x_i$ are global coordinate of $\mathbb{C}^5$.
My attempt (In the specific case above): For any point $x$ in $X$, there exists local complex coordinates $(z_1,\dots,z_4)$ on $\mathbb{CP^4}$ and holomorphic functions $f,g:\mathbb{C}^4\to\mathbb{C}$, such that locally $X$ is defined by $f=0$, and $$\dfrac{\Lambda}{F}=\dfrac{g(z)dz_1\wedge dz_2\wedge dz_3\wedge dz_4}{f(z)}.$$ By smoothness of $X$ and Implicit Function Theorem, there exists $i$ such that $\dfrac{\partial f}{\partial z_i}$ is nonzero at $x$. Since one can check easily that $$\dfrac{df}{f}\wedge(-1)^{i-1}\dfrac{g(z)dz_1\wedge\dots\wedge \widehat{dz_i}\wedge\dots\wedge dz_4}{\dfrac{\partial f}{\partial z_i}}=\dfrac{\Lambda}{F},$$ we have the explicit form : $$\text{Res}(\dfrac{\Lambda}{F})=\left.(-1)^{i-1}\dfrac{g(z)dz_1\wedge\dots\wedge \widehat{dz_i}\wedge\dots\wedge dz_4}{\dfrac{\partial f}{\partial z_i}}\right|_{f=0}.$$
I have tried to write down two coordinate chart, and compute the transformation rule, but the resultant is quite messy. Since $dz_1\wedge\dots\wedge \widehat{dz_i}\wedge\dots\wedge dz_4$ is of codimension 1, the expression will become a sum of 4 differential forms times cofactors of the Jacobian. On the other hand, the denominator also gives a sum of 4 terms, it seems hopeless to compute the transformation rule. I also think the restriction on $X$ plays a role, as it should helps reduce the computation on the denominator (using chain rule), but I am not sure at all. I have also tried to start from $\dfrac{df}{f}\wedge\rho=\dfrac{d\tilde{f}}{\tilde{f}}\wedge \tilde{\rho}$, for two coordinate charts, but I don't know how to proceed meaningfully. To be honest I have a feeling that I have missed something really obvious. Any help is appreciated!
Suppose $f$ defines $X$ on an open set $U$, and $g$ defines $X$ on an open set $V$. Then there is a non-vanishing holomorphic function $u$ on $U\cap V$ such that $f=ug$, where we can express $\omega$ as
$$\omega=\frac{df}{f}\wedge \rho=\frac{d(ug)}{ug}\wedge \rho=\frac{dg}{g}\wedge \rho+\frac{du}{u}\wedge \rho$$ Since $u$ non-vanishing, $\frac{du}{u}$ is holomorphic, so $\text{Res}(\frac{du}{u}\wedge \rho)=0$, so
$$\text{Res}(\frac{df}{f})=\text{Res}(\frac{dg}{g})=\rho,$$
which is independent of choice of coordinate.