Let $p(x,y)\in\mathbb{R}[x,y]$. Is it possible that $\partial_xp(x,y)=\partial_yp(x,y)=p(x,y)=0$ has a solution $(x,y)\in\mathbb{C}^2$ but no solutions $(x,y)\in\mathbb{R}^2$?
In other word, can you see if a real curve is "smooth" just by looking at the real points?
Of course, one could ask similar questions about arbitrary varieties over an arbitrary field $k$ (with $\mathbb{C}$ replaced by the algebraic closure $\bar{k}$): can a variety $X$ over $k$ be smooth at all $k$-points but have a singular $\bar{k}$-point, and if so, are there nice conditions (on $k$ or $X$) which make this impossible?
Sure. For a simple example, take $p(x,y)=(x^2+1)(y^2+1)$. This has no real points at all, but has singular complex points at $(\pm i,\pm i)$. (This example can immediately be generalized to any non-algebraically closed field, by letting $p(x,y)=q(x)q(y)$ where $q$ is a polynomial with no roots in the base field. Or, if you want to require $p$ to be squarefree, this works as long as the field is not separably closed. I don't know an irreducible example off the top of my head but I expect it is possible.)