Showing a function is discontinuous by taking a limit... For the function, 
$$
\begin{aligned}
f(x,y) = 
\begin{cases}
 \hfill 0 \hfill &,\mbox{ if }  x = y = 0 \\
 \hfill \frac{x^{2}y}{x^{4}+y^{2}} \hfill &, \mbox{ otherwise}  
\end{cases}
\end{aligned}
$$
Show it is discontinuous at $x=y=0$ by taking the limit $x \to 0$ along the line $y = {x^2}$. I don't even know what this means, I have a really unclear professor who mumbles through things half the time. Any assistance is appreciated.
EDIT:
What about the function, $$g(x,y) = \left\{ \matrix{
  0,x = y = 0 \hfill \cr 
  {{{x^2}y} \over {{x^4} + {y^2} + 1}},otherwise \hfill \cr}  \right\}$$
How do I know whether it's continuous at x=y=0 because I can't make a substitution like before...
 A: If a function of two variables should be continuos, that is $\lim_{(x,y) \to (a,b)} f(x,y) = f(a,b)$  independent on how you approach $a,b$. You could think of this like that you are drawing a circle around the point $(a,b)$ and independent from which point on the circle that you start from "walking" to the point $(a,b)$ you should "end" at the same value. 
So if we choose to start from the point $(x,x^2)$ we should end up at the value $0$ IF is continouis.
$$
f(x,x^2) = \frac{x^4}{x^4 + x^4} = \frac{1}{2} \not = 0, \ x \not = 0
$$
According to your second question, remember from early mathematics when you first learned about limits. I think it went something like this.
Lets say you have a one-dimension function $f(x)$ and you wanted to calculate the limit when $x \to a$, the first thing you would do is try to calculate $f(a)$. It $f(a)$ was definied, then you would simply say that $\lim_{x \to a} f(x) = f(a)$. If $f(a)$ was not definied, then you would have to do some "tricks". 
In this new case one simply sees that 
$$
f(0,0) = \frac{0}{1} = 0
$$
So this is no "dangerous" case, so we conlcude that $\lim_{(x,y) \to (0,0)} f(x,y) = f(0,0)$
