As regards $ f(x, y) = 4 + x^3 + y^3 - 3xy$, I computed that (0,0) is a saddle point, and (1,1) is a local minimum. So I'm not asking about this, and am asking only about sketching contours.
$1.$ Here, I'm referring to the middle sketch entitled "Why not this?" in red. How can one determine that the saddle point is more like a $+$ sign at (0,0), and not an $\times$? To wit, how do you deterine the position/orientation of the two lines which pass through the saddle point?
$2.$ I'm only able to sketch the leftmost with the calculated information, so how would you complete the sketch? I realise that a computer graphed the answer, but I want to sketch as much as possible.
Near the origin you have $f(x,y) \approx 4 - 3xy$, so the level curves there look roughly like the level curves of $xy$. (Multiplying by $-3$ and adding $4$ only changes the labelling of the level curves but not the actual shape of the curves.) The level curves of $xy$ are the $x$ and $y$ axes, together with hyperbolas $y=\text{constant}/x$.
The actual level curves of $f$ (when you include the higher order terms) may be bent as you move away from the origin, but the pair of level curves that go through the origin must be tangential to the $x$ and $y$ axes there (i.e., they must have the same direction as the corresponding level curves for the linearized function).
Another thing, which helps drawing the global picture, is that $f(y,x)=f(x,y)$, so the whole picture must be symmetric with respect to reflection in the line $y=x$.
And an "accident" which explains why the level set $f(x,y)=3$ is a straight line through the points $(0,-1)$ and $(-1,0)$: $$ f(x,y)-3 = x^3+y^3-3xy+1=(x+y+1)(x^2+y^2-xy-x-y+1) . $$