Proving a limit of a multivariable function does not exist 
Theorem: If $f(x,y)$ approaches two different values as $(x,y)\to (a,b)$ along two different paths in the domain of $f$,
  then $\lim_{(x,y)\to(a,b)}f(x)$ does not exist.

In class we had a function $f(x,y)$ with $(x,y) \to (0,0)$ and we showed that the limit of $f$ as $x \to 0$ was the same for all linear paths defined by $y=mx$.
In contrast, by making the substitution $y=ax^2$ we showed that the limit of $f$ as $x\to 0$ was different for parabolic paths, and therefore that the limit didn't exist.
My question is: 

How is this possible? Why can't we find some line of the form $y=mx$
  for any given parabolic path such that these paths both approach the
  limit in the same way? 
Why is it insufficient to prove that a limit exists just by
  substituting $y=mx$? If I remember calculus 2 correctly, we can pretty
  much approximate any function near $x=0$ with a linear function. So,
  what's going on?

 A: You can approximate not any  function, only differentiable functions. Not all functions are differentiable. 
Proving a limit at a point exists by just substituting y=mx means the function has a directional derivative at that point, along the direction of slope $m$.
A differentiable function has directional derivatives along all directions, but a function may have directional derivatives without being differentiable. That is the basis (in the more general context of normed spaces) of the difference between the Gâteaux derivative  and the Fréchet derivative.
A: Let we analyse what happens when dealing with one parameter functions. The domain of one parameter function is a line (or part of it) and it has only two directions (left and right). There you just need to see if the left-hand limit and right-hand limit are equal to the value of the function on a given point.
In two parameter functions the domain has a form of a plain, and so there are infinitely-many ways of different natures (line, parabola, spiral, ... even irregular curves) tending to a given point. To say that the limit exists, you need to have the same value in every direction when approaching the given point (equal to the value of the function on that point), otherways you'll have a wedge on the graph of the function respecting to the way approaching the point has a different value regarding the value of the function on that point.
A: This example might help you. Let $f:\mathbb R^ 2\to \mathbb R$ be defined in polar coordinates $r\ge0, \, \theta \in (0,2\pi]$ as $$f(r,\theta) = re^{1/\theta}.$$
In terms of these coordinates, approaching the origin through a line amounts to fixing $\theta$ and letting $r\to 0$. It is clear that, in any of those cases (with $\theta\in(0,2\pi]$) $f$ will approach $0$, but at different rates! 
This subtlety of $f$ is not captured by any straight line and is crucial to its (dis)continuity at the origin. Indeed, you can approach $\vec 0$ by a curve that passes through each line only once and along which $f$ is constant:
\begin{align} && f(r,\theta) &= C  \\ \iff&&  re^{1/\theta}&=C \\ \iff&& r &= Ce^{-1/\theta}.\end{align}
Hence, as $(r,\theta)$ approaches the origin along the curve $ (e^{-1/\theta},\theta)$ $f$ will take the value $1$, constantly.
Here's a contour plot of $f$:

