PART ONE:
I know how to find extremas by just using the gradient of a function $\nabla f(x,y)=0$
But I have been given a function alongside a constraint equation. And I am immediately thinking to evaluate this problem as a Lagrange multiplier problem but that is not a choice.
What is the theorem of finding absolute extremas with a function and a constraint?
i.e. $$f(x,y)= -x^2-y^2$$
$$x=y^2$$
$$C: 1 \le x\le 2$$
PART TWO:
If I am asked to find critical points and behavior with Lagrange multipliers and the equation has a different surface from the surface of the constraint, do I set them equal to one another to create a new constraint?
i.e. if $f(x,y)$ is a paraboloid and $g(x,y)$ is a flat plane, do I then say my constraint: $f(x,y) = g(x,y)$?
PART THREE:
A third question is, how would the second derivative test fit into Lagrange Multipliers? Or is it unnecessary? I have not used the second derivative test in a Lagrange multiplier problem yet but I have an idea that it is possibly a way to solve for maximum and minimum points in this type of problem.
Part I: You are given a function $f:\>{\mathbb R}^2\to{\mathbb R}$ (not a "surface") as well as a feasible set $$S:=\bigl\{(x,y)\,\bigm|\,x=y^2, \ 1\leq x\leq2\bigr\}\ .$$ The set $S$ consists of two arcs on the parabola $x=y^2$. These arcs can be presented in the form $$t\mapsto (x,y):=(t^2,\pm t) \qquad(1\leq t\leq\sqrt{2})\ .$$ This allows to avoid Lagrange multipliers. Instead we look at the pullback $$\hat f(t):=f\bigl(x(t),y(t)\bigr)=-t^4-t^2\qquad(1\leq t\leq\sqrt{2})\ .$$ It is obvious that $\hat f$ is maximal when $t=1$, leading to the two points $(1,\pm1)$, and $\hat f$ is minimal when $t=\sqrt{2}$, leading to the other two endpoints $(2,\pm\sqrt{2})$ of the two arcs in question.
Part II: I suggest you remove this part from your question, as it does not make any sense.
Part III: Note that a second derivative test could only tell you whether a conditionally stationary point found by Lagrange's method is a local conditional minimum or maximum of $f$. There is such a test; it involves second derivatives of the constraint as well. I have never used it. For determining the global extrema of $f$ on a feasible set $S$ it is absolutely indispensable to create a candidate list encompassing all points found with Lagrange multipliers, then the analysis of edges (if any), and the vertices or endpoints of arcs, as in the above example.