Specific path in a gradient field of a saddle I think this is a classic question but I could not find a useful reference to understand it so I post it.
Take the classic saddle $z=x^2-y^2$. When you sit over it in $(0,0)$ you see in front of you that the function rises and your legs are over a parabola which is the less wide one.
Now take a more complicated saddle-like function such as $z=(3x+3y)^2-(4x+y)^2$. In this way the saddle isn't symmetric and I want to determine the curve which has the same role of the "less wide parabola" of the first case (so where to put my legs comfortably in a certain sense)
I thought that a possible solution could be to calculate the gradient field (which is immediate since only derivations are needed) and then determine the path which goes through the "steepest descent" through the gradients. So this path is the curve I'm looking for (the equivalent of the "less wide parabola").
Is the approach to my problem correct? 
How do I calculate analytically the steepest descent?
Indeed I need a procedure that can be applied to more complicated functions than the two mentioned. 
Thank you!
 A: First calculate the Hessian matrix at your saddle point $(x,y) = (x_0,y_0)$,
$$
H(f)(x_0,y_0) = \left(\begin{array}{cc}
f_{xx}(x_0,y_0) & f_{xy}(x_0,y_0) \\
f_{yx}(x_0,y_0) & f_{yy}(x_0,y_0)
\end{array}\right).
$$
If the saddle point is nondegenerate this will have one positive and one negative eigenvalue.  The eigenvector associated with the negative eigenvalue is tangent to the path of steepest descent from the saddle and the eigenvector associated with the positive eigenvalue is tangent to the path of steepest ascent.
For your example
$$
g(x,y) = (3x+3y)^2-(4x+y)^2
$$
we have
$$
H(g)(0,0) = \left(
\begin{array}{cc}
 -14 & 10 \\
 10 & 16
\end{array}
\right),
$$
which has eigenvalues $\lambda_{\pm} = 1 \pm 5\sqrt{13}$ with associated eigenvectors 
$$
\mathbf{v}_{\pm} = \left(\frac{1\pm 5\sqrt{13}}{10} - \frac{8}{5},1\right).
$$
So the path of steepest ascent is tangent to $\mathbf{v}_+$ and the path of steepest descent is tangent to $\mathbf{v}_-$.
Below is a plot showing the straight line in the direction of $\mathbf{v}_+$ in $\color{red} {\text{red}}$ and the straight line in the direction of $\mathbf{v}_-$ in $\color{blue} {\text{blue}}$ on the graph of $g(x,y)$.

If your saddle is degenerate then you will need to look at higher-order coefficients in the Taylor expansion of $f$ centered at the saddle.
