Can delta depend on $x$ and not just $\epsilon$ ? Plus, example. In the definition of a limit, can $\delta$ depend on a variable as well? I don't see anything about this in my book. 
Also, I want to find the limit of  $xy \frac{x^2 - y^2}{x^2 + y^2}$ as $(x,y)$ approach $(0,0)$, if it exists at all. If either $x$ or $y$ are zero, the whole thing is zero, so my guess is $0$. Then, by the definition, we see that $$|xy \frac{x^2 - y^2}{x^2 + y^2}|  = |x||y||\frac{x^2 - y^2}{x^2 + y^2}| \le |x|y| \le |x|\sqrt{x^2 + y^2}$$ and so I pick $\delta = \epsilon /|x|$ because then $$|x|\sqrt{x^2 + y^2} \le |x| \delta = \epsilon$$
But, this being my first independent attempt at finding a limit for a multivariable function, I probably made a mistake somewhere. 
Also, my text book asks for the existence of a limit before the limit itself, i.e. we're supposed to determine existence before the numerical value of the limit itself. I don't think I do this, so how would one do that? 
 A: To your first question: It depends whether you want to prove continuity or uniform continuity:


*

*For the prove of continuity $\delta$ may depend on $\epsilon$ the argument $x$ where the function shall be continuous.

*For the prove of uniform continuity $\delta$ can only depend on $\epsilon$.


Note that uniform continuity is always a property of the whole function while continuity can be a property of the whole function and and also be a property that the function only has at certain points.
Note 1: The term $xy\frac{x^2-y^2}{x^2+y^2}$ has no value at $(x,y)=0$ because you would devide by zero there which is forbidden.
Note 2: You investigate the limit at $(\tilde x, \tilde y)=(0,0)$ and thus your $x$ in the formula $\delta=\frac{\epsilon}{|x|}$ is zero. Because you cannot divide by zero you have to take another $\delta$.
Hint: You can prove $$\|0 - xy\frac{x^2-y^2}{x^2+y^2}\| \le x^2+y^2 = \|(x,y)\|^2$$
Your task is now: Given any $\epsilon > 0$, find an $\delta > 0$ such that $$\|(x,y)-(0,0)\| < \delta \Rightarrow \|0 - xy\frac{x^2-y^2}{x^2+y^2}\| < \epsilon$$
I guess you will find the necessary $\delta>0$ easily because of the work is already done... ;-)
