How to Complete Sketch of a function of two variables $ f(x, y) = 3x - x^3 - 2y^2 + y^4$ ? [Stewart P930 Question 14.7.4] For $ f(x, y) = 3x - x^3 - 2y^2 +  y^4$ $\implies$
$\partial_x f = 3 - 3x^2, \partial_y f = -4y + 4y^3$. Set both equations to 0 $\implies x = \pm $1 and $y = 0, \pm 1$.

$1.$ To determine the critical points, how does one determine which combinations of x and y to form/pair up? The answer is 6 stationary points: for each of the two x values, each of the 3 y-values constitute a statioanary point? I see that $\partial_x f$ is independent of y and $\partial_x f$ of x.

I'm not asking how to compute: (1,0) is a local maximum point,
$(1,\pm 1)$ are saddle points,
$(-1,0)$ is a saddle point,
$(1,\pm 1)$ is a local minimum 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.
Moreover, how do you determine that the sketch has an upper/lower/left/right bound (as signaled by my red arrows)?


 A: If you're doing this by hand, also pay attention to the asymptotics (when $x \to \infty, y \to \infty$. For large $y$, $y^4$ dominates, and for small $y$ and large $x$, $-x^3$ dominates. However, I doubt that you can get a very good sketch, even after finding critical points, without the aid of computer.
A: Note also the symmetry $y \to -y$ (reflection across the vertical axis).
In addition, note that the slope of your curve is 
$$ \dfrac{dy}{dx} = - \dfrac{\partial_x f}{\partial_y f}$$
so you see the regions where these are positive and negative.
In the left half-plane, your function $f$ goes to $+\infty$ as $|x|+|y| \to -\infty$ so the curves stay bounded there.  If you follow a curve $f(x,y)=c$ starting, say,
on the $x$ axis to the left of $x=-1$ as it goes into the upper half-plane, 
it starts out vertical, then bends to the left (negative slope in the region
$x < -1$, $0 < y < 1$).  It must hit $y=1$ at some point, again going vertically,
then goes up and right until it hits $x =-1$ (going horizontally), then down and to the right.  What happens next depends on whether $c$ is above or below 
$f(1,1) = 1$.  If it's below, it hits $y=1$ (vertically), then bends to the left and continues until it hits $y=0$ (again vertically).  By symmetry, the rest of the curve is the reflection of this part across the $y$ axis. 
