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)?
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.