Why does $y$ go to $0$ faster than $x$? Let $\frac{x^2}{y}\to c\neq 0$ as $(x,y)\to (0,0)^+$. Why does this mean that $y$ goes faster to $0$ as $x$ does?
For values of x and y that are small enough, we surely do not have $x^2/y=1$ so they cannot be identical, okay. Moreover, if the quotient would tend to 0, this would mean that $x$ would be faster decaying to 0 than y. 
 A: Here is kind of out of the box and probably too complicated view on the matter : 
You may want to think of $x$ and $y$ as positive, decreasing, differentiably-continious  functions of the variable $t$ for which $x(t),y(t) \xrightarrow[t \to \infty]{} 0$. Now by L'Hospital's Rule you will obtain that
$$
\lim\limits_{t\to\infty}\frac{x^2(t)}{y(t)} = \lim\limits_{t\to\infty}\frac{2x(t)x'(t)}{y'(t)} = C \neq 0. 
$$
This means that $\lim\limits_{t\to\infty}\frac{x'(t)}{y'(t)} = \lim\limits_{t\to\infty}\frac{C}{x(t)} = \infty$ and thus there exists $T$ for which for $t > T$ you have $0 > x'(t) > y'(t)$. We actually obtained that that the "speed" of decreasing of $y$ is greater than that of $x$.
A: So first of all let me make the statement a little more precise. You are looking at the function
\begin{eqnarray}
f:\mathbb{R}^2\setminus \{0\} &\to & \mathbb{R}\\
(x,y) &\mapsto & \dfrac{x^2}{y}.
\end{eqnarray}
Now what does $\lim_{(x,y)\to(0,0)^+}f(x,y)$ actually mean? It means you take any sequence of $(x_n,y_n)$ which approaches zero s.t. $x_n,y_n>0$ for all $n$ and you evaluate $f$ on this sequence. So what you actually have is $\lim_{n\to \infty} f(x_n,y_n)$. Now what does it mean that $f(x_n,y_n)\to c \neq 0$. This is best illustrated by an example: Let $x_n:=\frac{1}{n}$ and let $y_n:=\frac{x_n^2}{c}$ now you have:
$$\lim\limits_{n\to \infty} f(x_n,y_n)=\lim\limits_{n\to \infty}\dfrac{\dfrac{1}{n^2}}{\dfrac{1}{n^2}}\cdot c=c $$.
In this example you can allready see that $y_n$ approaches zero faster then $x_n$ since $y_n=1/(c\cdot n^2)$ and $x_n$ is "only" $1/n$. Now lets make that more precise: assume that $x_n$ and $y_n$ approach zero "at the same speed" i.e. $\lim_{n\to\infty} \frac{x_n}{y_n}=r$ for some real number $r \in \mathbb{R}$. Then using the algebraic limit theorem we would have:
$$\lim\limits_{n\to \infty}f(x_n,y_n)=\lim\limits_{n\to \infty}x_n\cdot \lim\limits_{n\to \infty} \dfrac{x_n}{y_n}=0\cdot r=0. $$
But we excluded this case by assumption. So $\lim\limits_{n\to \infty} \frac{x_n}{y_n}=r$ can't be true. 
Let me wrap it up a little: we want to show that under the condition that $\lim\limits_{n\to \infty}f(x_n,y_n)=c\neq 0$ and $x_n,y_n>0$ it must hold that $\lim\limits_{n\to \infty}\frac{x_n}{y_n}=\infty$ or equivalently $\lim\limits_{n\to \infty} \frac{y_n}{x_n}=0$. Now since $x_n,y_n> 0$ we have that $\frac{x_n}{y_n}>0$. Suppose that $\frac{x_n}{y_n}$ is bounded by some $C$ then we woud have that $ 0<f(x_n,y_n)\leq x_n\cdot C$ and taking the limit we would have that $\lim\limits_{n\to \infty} f(x_n,y_n)=0$ contradicting the assumption. The existence of the limit $\lim\limits_{n\to \infty} f(x_n,y_n)=c$ ensures that indeed $\frac{x_n}{y_n}$ approaches infinity showing what we wanted to show.      
