Nonsingularity of $xy-z^2$ I am trying to show that $Z(xy-z^2)$ in $\mathbb{P}^2$ is nonsingular.
My idea is considering the first standard chart $\mathbb{A}^2\hookrightarrow\mathbb{P}^2$ given by $(x,y)\mapsto [x:y:1]$. Intersecting $Z(xy-z^2)$ with the chart gives $Z(xy-1)$. The complement of the first standard chart is a copy of $\mathbb{P}^1$, $[x:y:0]$. And $Z(xy-z^2)\cap\mathbb{P}^1_{x:y}=[1:0:0]\cup[0:1:0]$.
As I learned before, $Z(xy-z^2)$ is isomorphic to $\mathbb{P}^1$... The hint says do it for the first affine chart, then check the points not in the chart. I wonder how should I continue the proof.

If you want to understand the tangent space at a given point, you probably want more than just that point; you want an open neighbourhood.
So instead of looking just at $[1:0:0]$ look at the open neighbourhood $\{x = 1\}$ which gives you the equation $y - z^2$.
The three standard open sets give you three equations: $y - z^2$, $x - z^2$, $xy - 1$.
If you take partial derivatives you get three Jacobians: $(1, -2z), (1, -2z), (y, x)$. Now check that $J = (0, 0)$ only if $z = 1$ and $x, y = 0$.
