Suppose $X$ is a variety, $U\subset X$ is an open subvariety, and $p\in U$. As far as I understand, the local rings at $p$ are isomorphic: $\mathcal{O}_{X,p}\simeq \mathcal{O}_{U,p}$, and therefore so are the cotangent spaces, and thus $p$ is a singular point of $U$ if and only if it is a singular point of $X$.
However, this does not seem to be true in the following example: let $X = \{ y^2=z^2+xz\}\subset \mathbb{P}^2$ and $p=(1:0:-0.5)$. Then $U = X \cap \{(x:y:1)\}$ is an open neighborhood of $p$, and $U\simeq \{y^2=1+x\}\subset \mathbb{A}^2$. Calculating the gradient of $f(x,y)=y^2-(1+x)$, we see that $U$ has no singular points. On the other hand, if we take the open neighborhood $V= X \cap \{(1:y:z)\}\simeq\{y^2=z^2+z\}\subset \mathbb{A}^2$, a similar calculation shows that $p$ is singular.
Where did I go wrong?
Evaluating $y^2-z^2-xz$ at $P$ gives $0-\frac{1}{4}+\frac{1}{2}=\frac{1}{4}\neq 0$, so $p\notin X$. So you cannot talk about whether $p$ is a singular point of $X$, as $p\notin X$.