Open set of the cuspidal curve is not principal

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I'm struggling with this problem:

Let $k$ a field and $X=\operatorname{Spec} k[x,y]\big/(x^3-y^2)$ and $U=X\setminus\{(x-1,y-1)\}$. Show that $U$ is not a principal open set in $X$.

I tried to show that we can't find a polynomial in this ring with only $(1,1)$ as zero, but I found problems related to $k$ is any field with any characteristic and not necessarily algebraically closed. I tried also to use the same arguments in $k[t^2,t^3]$ but they lead nowhere.

Thanks in advance to those who can help me.

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Over a field of positive characteristic, $U$ is a principal affine open. Over a field of characteristic zero, $U$ is not a principal affine open.

The reasoning comes from the ability to write down a function $f\in k[t^2,t^3]$ vanishing only at $t=1$: such a function must be a power of $(t-1)$, but written as a polynomial in $t^2$ and $t^3$. When the characteristic is zero, $(t-1)^n$ always has a term $\pm nt$ which is never zero and any polynomial with a nonzero $t$ term cannot be written as a polynomial in $t^2$ and $t^3$. When the characteristic is positive, $t^p$ can be written as a product of $t^2$ and $t^3$, so $t^p-1=(t-1)^p$ is a function vanishing only at $t=1$ and therefore showing that $U$ is a principal affine open subset.