Let $f(x,y)=(x^2)/(x^2+y^2)$, show that the domain for this function is all points $(x,y)$ except $(0,0)$ 
Let $f(x,y) = \frac {x^2} {x^2 + y^2}$. The domain for this function is all points $(x,y)$ except $(0,0)$ and it is continuous for all $(x,y)$ in its domain. Can $f(x,y)$ be made continuous at $(0,0)$ by defining $f(0,0)$ to be a specific value? Explain! Draw the level curves for this function for function values: $0$, $0.5$ & $1$.

I don't understand this question and I don't know also how start .
I haven't any program to draw the curve so is there free website can draw that curves?
 A: Hint. You may observe that
$$
f(x,y)=\frac{x^2}{x^2+y^2}
$$ gives, as $x \to 0$,
$$
f(x,0)=\frac{x^2}{x^2} \to 1
$$ and, as $x \to 0$,
$$
f(0,x)=\frac{0^2}{0^2+x^2} \to 0.
$$

Thus it is not possible to make $f$ continuous at $(0,0)$.

A: For the function to be continuous , it is necessary that the limit, as you approach the potential trouble point $(0,0)$in any direction, is the same. But, if you approach $(0,0)$, along the curves $(y,ny)$, the limit will not be the same $$\frac {n^2y^2}{y^2+n^2y^2} \rightarrow \frac {n}{n+1}$$, and the value
$\frac {n}{n+1}$ is different for each $n$, so the function cannot be made continuous at $(0,0)$.
On the other question, the level curves for $1, 0.5$, this means drawing the graphs of the constant functions $f(x,y)=1, f(x,y)=0.5$:
$$\frac {x^2}{x^2+y^2}=1 \rightarrow x^2= x^2+y^2$$ and $$ \frac {x^2}{x^2+y^2}=0.5 $$ respectively.
Wolfram  http://www.wolframalpha.com/input/?i=plot+3d+x%5E2%2F%28x%5E2%2By%5E2%29  will plot it for you for free, if you enter: Plot 3d x^2/(x^2+y^2) , but you must deposit 10,000 USD in my account first ;).
A: The level sets are going to be easier to draw than you may think. First, solve $f(x,y)=\alpha$ for $\alpha\in\{0,0.5,1\}$ to see what they will look like. In fact, doing so should help you to answer the rest of the question!
For example, the following are equivalent for all $(x,y)$ in the domain of $f$: $$f(x,y)=1$$ $$x^2=x^2+y^2$$ $$0=y^2$$ $$0=y$$
Hence the set of all $(x,y)$ such that $f(x,y)=1$ is the set of all points of the form $(x,0)$ in the domain of $x$--that is, it is just $x$-axis with the origin removed!
A: They basically ask you to prove or disprove
$$
\lim_{(x,y)\to(0,0)} \frac{x^2}{x^2+y^2} = 0
$$
Hint: Examine the limit on the path $x=y$.
For the plot, you can use WolframAlpha.
