Hey Elijah, the answer to your question is quite simple, elementary and explicit! You don't need to read up on anything fancy. Here it goes:

**Fact:** The canonical divisor of a smooth affine hypersurface is **zero**! In particular the canonical divisor of your curve $f(x,y)$ is $0$ since as you have mentioned the curve has a smooth projectivisation (so the curve itself must be smooth).

**Proof:** Let $X\subset\mathbb{A}^2$ be the affine curve defined by $f(x,y)$ which we are assuming to be smooth. Define the open sets $U_1,U_2$ in the plane by $\frac{df}{dx}\neq{0}$ and $\frac{df}{dy}\neq{0}$. Then $y$ and $x$ are local parameters in $U_1$ and $U_2$ respectively and the forms $dy$ and $dx$ are the basis of $\Omega^{1}[U_1]$ over $k[U_1]$ (respectively $\Omega^{1}[U_2]$ over $k[U_2]$). However, let us choose more convenient basis like $\omega_1=-\frac{dy}{df/dx}$ and $\omega_2=\frac{dx}{df/dy}$ on $U_1$ and $U_2$ respectively. This is permissible since the denominators don't vanish on the respective open sets. Now note that on $U_1\cap{U_2}$ both the forms are equal since $\frac{df}{dx}dx+\frac{df}{dy}dy=0$, therefore they patch to give a form $\omega$ that is regular and everywhere nonzero on $U$, so that $div\ \omega=0$ in $U$. In other words the canonical divisor is zero.

Note: This works analogously for any smooth affine hypersurface.

P.S.: Quoting an exact sequence is not a substitute for making even one small and simple calculation. Hope this motivates you for more algebraic geometry!