= Solution
The <second fundamental form> records the normal component of second-order bending of the surface. For a surface curve $\gamma(t)=\sigma(u(t),v(t))$, the terms containing $u''$ and $v''$ are tangent and therefore
$$
\gamma''\cdot\boldsymbol N=L(u')^2+2Mu'v'+N(v')^2.
$$
Thus on a unit tangent direction it gives the <normal curvature>; for a general coordinate direction the <normal curvature> is the ratio of the second to the <first fundamental form>. Equivalently the quadratic part of displacement out of the <tangent plane> is one half of this form. Reversing the chosen <unit normal> changes its sign.
Now suppose $L=M=N=0$. Differentiate $\boldsymbol N\cdot\sigma_u=\boldsymbol N\cdot\sigma_v=0$. This gives
$$
\boldsymbol N_u\cdot\sigma_u=-L=0,\quad\boldsymbol N_u\cdot\sigma_v=-M=0,\qquad\boldsymbol N_v\cdot\sigma_u=-M=0,\quad\boldsymbol N_v\cdot\sigma_v=-N=0.
$$
Also differentiation of $\boldsymbol N\cdot\boldsymbol N=1$ gives $\boldsymbol N_u\cdot\boldsymbol N=\boldsymbol N_v\cdot\boldsymbol N=0$. The parametrization is regular, so $\sigma_u,\sigma_v,\boldsymbol N$ form a <basis> of three-dimensional space. Both normal derivatives are orthogonal to this <basis> and hence vanish:
$$
\boxed{\boldsymbol N_u=\boldsymbol N_v=0.}
$$
The parameter ball is <connected>; integrating these derivatives along a straight segment between parameter points shows that $\boldsymbol N=\boldsymbol N_0$ is constant. Finally both derivatives of $\sigma\cdot\boldsymbol N_0$ vanish, because $\sigma_u,\sigma_v$ are tangent. Thus the <zero second fundamental form implies planar image> statement is
$$
\boxed{U\subset\{\boldsymbol x:\boldsymbol x\cdot\boldsymbol N_0=C\}.}
$$
The PDF has $\boldsymbol N_v$ in the second derivative equality; the apparent $\boldsymbol N_t$ in the TeX is not a distinct coordinate.
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