Easy way to determine the sign of Gaussian curvature without explicit computation. I know how to compute Gaussian curvature via the first and second fundamental form. i.e., 
\begin{align}
K = \frac{eg-f^2}{EG-F^2} 
\end{align}
In this computation one usually set $n = \frac{X_u\times X_v}{||X_u\times X_v||}$
Without detail computation through the formula above, is there an easy way to determine the sign of Gaussian curvature? 
For example, with some computation, I know the Gaussian curvature of this type of surface $ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}=1$ is $K = \frac{1}{a^2 b^2 c^2 (\frac{x^2}{a^4} + \frac{y^2}{b^4} + \frac{z^2}{c^4} )^2}$ and for this type of surface $\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} =1$, $  K(p) = -\frac{1}{a^2 b^2 c^2 (\frac{x^2}{a^4} + \frac{y^2}{b^4} + \frac{z^2}{c^4} )^2}$. 
To figure out the sign it took me lots of time.  I want to know the following : Is there an easy way to figure out whether the curvature is positive or negative rather than computing explicitly?

I suppose you are considering surfaces (ie 2-dimensional) embedded into $\mathbb{R}^3$. You can try to formalize the intuition that montains and valleys have positive curvature and saddle points have negative curvature. 
Compute the tangent plane of the surface at a point. If in a small neighbourhood of the point, the surface lies entirely on one side of the tangent plane, the Gaussian curvature is positive at that point. If the surface intersects the tangent plane, the Gaussian curvature is negative.
