This is about the definition of nonsingularity of a plane curve defined over arbitrary field $k$.
By convention a plane curve over $k$ is defined by some $f(x,y)\in k[x,y]$ but the whole curve lives $V_\bar{k}(f)\subset \mathbb{A}_\bar{k}^2$, where the $k$-rational points are $V_k(f)=V_\bar{k}(f)\cap \mathbb{A}_k^2$, which we assumed to be non-empty in this case.
There is an ambiguity about the definition of curve here, $V_k(f)$ or $V_\bar{k}(f)$.
As a variety we must have at least $V_k(f)$ is irreducible but $f$ irreducible in $k[x,y]$ is not enough so it is natural to assume $f$ is irreducible in $\bar{k}[x,y]$, i.e. $f$ is geometrically irreducible.
Now to define the nonsingularity of the plane curve $f$, most of the textbooks just say check the singularity over the curve, where $p$ is a singular point if $f(p)=f_x(p)=f_y(p)=0$.
Clearly checking over $V_\bar{k}(f)$ will probably be what we want, but is it enough to check over $V_k(f)$?
Thus I want to either prove
Given a geometrically irreducible non-constant polynomial $f\in k[x,y]$, with at least one $k$-rational point, if the variety $V_\bar{k}(f)$ is nonsingular on all $k$-rational points, then it is nonsingular over all points.
or find a counter example: Find a geometrically irreducible non-constant polynomial $f\in k[x,y]$, with at least one $k$-rational point, nonsingular on $V_k(f)$ but singular on $V_\bar{k}(f)\backslash V_k(f)$.
Consider the polynomial $$f(x,y)=x(x^2-2)^2-y^2$$ over the rationals. Its gradient $$\nabla f(x,y)=(5 x^4-12 x^2+4,-2 y) $$ vanishes at $(\pm\sqrt{2/5},0)$ and $(\pm\sqrt2,0)$. Of these the latter two are on the zero locus of $f$, so $V(f)$ is non-singular at all the rational points, but has singularities in a quadratic extension $\Bbb{Q}(\sqrt2)$.
Furthermore, $f(1,1)=0$, so there are rational points in $V(f)$. Also, $f(x,y)$ is irreducible in $\Bbb{C}(x)[y]$. Apply Eisenstein's criterion with the prime $p=x$.