Let $U\subset\mathbb R^2$ be open and connected, and let $\ f:U\rightarrow\mathbb R^3$ be the parametrization of a regular surface. Prove that $\textbf{II}_p\equiv0$ on $M:=f(U)$ if and only if $M$ is contained in a plane
(where, $\textbf{II}_p$ is the second fundamental form)
$\textbf{My attempt}$
$''\Rightarrow''$
Let $\textbf{II}_p\equiv0$, then
$f_{uu}\cdot\ n=0$$\qquad$$f_{uv}\cdot\ n=0$$\qquad$$f_{vv}\cdot\ n=0$$\qquad$ since $n$ is nonzero vector
$f_{uu}=0$ this implies $\ f_u$ is constant (the other cases analogue)
so $f(u,v)=su+tv$ ($s,t$ constants) and this is a subset of a plane ?
From your proof, you only get $f_{uu}, f_{uv}, f_{vv}$ are orthogonal to $\vec n$, and $\vec n$ is a priori not a constant vector in $\mathbb R^3$. Moreover, it is impossible to obtain $f_{uv} = f_{uu} = f_{vv}=0$, for example, take $g: U \to U$ be any diffeomorphism and $f = i\circ g$, where $i(x, y) = (x, y, 0)$.
A standard proof is to consider geodesics. Try to show that a geodesic in $f(U)$ is also a geodesic in $\mathbb R^3$. What is the geodesic in $\mathbb R^3$?