Computing the genus of $y^2z=x^3$ using differentials

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In Vakil's Foundations of Algebraic Geometry, example 21.2.7 he takes the curve $y^2=x^3-x$ in $\Bbb{A}_k^2$ (with $\text{char}(k)\neq 2$) and defining $A:=k[x,y]/(y^2-x^3+x)$ he shows that $\widetilde{\Omega}_{A|k}$ is an invertible sheaf.

He writes $2ydy=3x^2dx$ and notices that $\Omega_{A|k}$ is generated by $dx$ when $y\neq 0$ and generated by $dy$ when $3x^2-1\neq 0$. Since $\{y\neq 0\}$ and $\{3x^2-1\neq 0\}$ cover the whole curve, that's enough.

It's nice to look at things this way and I wonder if I could do something like that for different purposes.

For example, could I find the geometric genus $p_g=\dim H^0(X,\Omega_{X|k})$ for a given projective curve $X\subset \Bbb{P}^2$?

I tried to take the curve $y^2z=x^3$ in $\Bbb{P}^2_{(x:y:z)}$. Taking each affine chart, we have: \begin{align*} z=1&:\,\,2y=3x^2dx\\ y=1&:\,\,dz=3x^2dx\\ x=1&:\,\,2yzdy+y^2dz=0 \end{align*}

I was expecting to find $p_g=0$ (since a singular plane cubic is rational), but I don't know how to see that by the equations above.