Prove a surface with $K<0$ has no umbilic points Also, prove an umbilic point on a surface with $K\leq0$ is a planar point
Ok, here’s my attempt.
An umbilic point is a point whose principal curvatures are equal.
We know Gaussian curvature can be expressed as the product of principal curvatures $k_1,k_2$. 
If $k_1=k_2$ then $K=0$ or $K>0$.
We conclude if $K<0$ a surface has no umbilic points
Then, if $K\leq0$ and $K<0$ has no umbilic points, we must have $K=0$ which implies our umbilic point lies on a plane and thus is planar
 A: Take for example a hyperbolic paraboloid with cartesian equation $z=\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2}$. It has negative gaussian curvature in each of its points but the sectional curvature is not zero for every curve that passes through a point $p$ (you can see that if you graph the intersection of the surface with the planes $x=0$ and $y=0$. So basically you cannot have principal curvatures with different signs because that would always give you positive gaussian curvature. That means that both principal curvatures must have different signs, so the only way you can have $K\geq 0$ is if both principal curvatures are zero. Which is pretty much what you said, which looks right to me.
A: It is elementary algebra with real scalars $k1$ and $k2$ representing principal curvatures.
If product $K$ of two real numbers is negative, then the factors $k1,k2$ should be of opposite sign.
An umbilic point should have same sign (and magnitude) of curvature for the two factors. It should necessarily have $ K>0$.
