If the ground field has characteristic $p$ then every line through the origin is a tangent line to the curve $y = x^{p+1}$ I tried simple example $F_{2}$ and $y=x^3$. The there are only 2 points (0,0) and (1,1). Then how to prove that every line through the origin is a tangent line?

Suppose $L$ is a line through the origin, so $L$ is given by the equation $y=mx$ for some $m$. The intersection of $L$ with the curve $y=x^{p+1}$ is $mx=x^{p+1}$ which is equivalent to $x(x^p - m)=0$, or $x(x-\sqrt[p]{m})^p=0$ because we are in characteristic p. By looking at the zeros of this equation, the intersection points correspond to $x=0$ (the origin) or $x=\sqrt[p]{m}$ with multiplicity $p>1$, so that is why the line $L$ is tangent to the curve.
Ref:Tangent lines to a curve passing through a given point
