= Solution
The reduction has equation $F=x^3+y^3=0$. At the origin both first <partial derivatives> vanish, so the point is singular, also when $p=3$. A lift reducing to the origin would require $x,y\in p\mathbb Z_p$. Its two cubes would then lie in $p^3\mathbb Z_p$, whereas the right side of the original equation has <valuation> one. This is impossible.
For an example which does lift, choose
$$
\boxed{\mathcal D:x^3+y^3=p^3.}
$$
It has the same reduction and the <rational point> $(p,0)$ reduces to the origin. Over $\mathbb Q_p$, which has <characteristic> zero, a singular affine point would require $x=y=0$, which does not lie on this curve. Its <projective closure> is also smooth: the simultaneous <partial derivatives> of $X^3+Y^3-p^3Z^3$ cannot vanish at a projective point. Thus this is a nonsingular curve for every <prime number>[prime] $p$, including three. A <singular point> in the reduction can therefore fail to lift or can lift, depending on the original equation.
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