Proof verification: non-continuity of $f(x,y)=\frac{2xy}{x^2+y^2}, f(0,0)=0$ at $(0,0)$ 



I am trying to prove that the following function is discontinous at $(x,y)=(0,0)$. For this, I'm using the definition provided by Courant in "Differential and Integral Calculus, Vol. 1" given above:
$$f(x,y)= \begin{cases} 
      \frac{2xy}{x^2+y^2} & (x,y) \neq (0,0) \\
      \quad  0 & (x,y) = (0,0)
   \end{cases}$$
So I took the definition:
$$\left| \frac{2xy}{x^2+y^2}\right| < \epsilon$$
Now we know that:

$$(x-y)^2 \geq 0 \quad \quad x^2+y^2 - 2xy \geq 0 \quad\quad 1 \geq \frac{2xy}{x^2 + y^2} \geq -1 \tag{$\star$}$$
$$0<a<b \iff 1> \frac{a}{b} > \frac{a}{b+1}\tag{$\star \star $}$$

Now let $\epsilon_0 = \frac{1}{n}$, then we have $x_0,y_0$ such that:
$$\left| \frac{2x_0 y_0}{x_0^2+y_0^2}\right| < \frac{1}{n}$$
Then we have a $\delta_0$ and $x_0^2+y_0^2< \delta_0^2$.

Now let $\epsilon_1=\frac{1}{n+1}$, then we have $x_1,y_1$ such that
$$\left| \frac{2x_1 y_1}{x_1^2+y_1^2}\right| < \frac{1}{n+1}$$
Then we have a $\delta_1$ and $x_1^2+y_1^2< \delta_1^2$.
As:
$$\frac{1}{n}>\frac{1}{n+1}$$
Then: 
$$1>\left| \frac{2x_0 y_0}{x_0^2+y_0^2}\right| > \left| \frac{2x_1 y_1}{x_1^2+y_1^2}\right| >0  $$
Now using $(\star)$ and $(\star \star)$, we can (?) deduce:
$$x_0^2 + y_0^2 < x_1^2 + y_1^2$$
Which implies that $\delta_0 < \delta_1 $. With this, we see that as $n\to \infty$, $\delta \to \infty$ and it implies that the function is not continuous at $(x,y)=(0,0)$. Is my proof correct? I guess that the only step that seems a little wishy-washy is where I marked with (?).
 A: 
So I took the definition: 
$$
\left| \frac{2xy}{x^2+y^2}\right| < \epsilon
$$

(A proof consists of complete logical sentences. This sentence is not complete. Proper quantifiers are missing.)

Now we know that:
$$(x-y)^2 \geq 0 \quad \quad x^2+y^2 - 2xy \geq 0 \quad\quad 1 \geq \frac{2xy}{x^2 + y^2} \geq -1 \tag{$\star$}$$
$$a<b \iff 1> \frac{a}{b} > \frac{a}{b+1}\tag{$\star \star $}$$

(Again, quantifiers are missing and the second statement is incorrect.)

Now let $\epsilon_0 = \frac{1}{n}$, (What is $n$ ? No definition found.) then we have $x_0,y_0$ such that:
  $$
\left| \frac{2x_0 y_0}{x_0^2+y_0^2}\right| < \frac{1}{n}
$$
  Then we have a $\delta_0$ and $x_0^2+y_0^2< \delta_0^2$.
(Very confusing. "Then" means a logic implication. Why the first "then"?
   How do you have the second "then"? And again, quantifiers are missing. Up to this point, it is unclear what you are trying to show.)



Now let $\epsilon_1=\frac{1}{n+1}$, then we have $x_1,y_1$ such that
$$ 
\left| \frac{2x_1 y_1}{x_1^2+y_1^2}\right| < \frac{1}{n+1}$$
Then we have a $\delta_1$ and $x_1^2+y_1^2< \delta_1^2$.
As:
$$\frac{1}{n}>\frac{1}{n+1}$$
Then: 
  $$
1>\left| \frac{2x_0 y_0}{x_0^2+y_0^2}\right| > \left| \frac{2x_1 y_1}{x_1^2+y_1^2}\right| >0 
$$
  Now using $(\star)$ and $(\star \star)$, we can (?) deduce:
  $$
x_0^2 + y_0^2 < x_1^2 + y_1^2
$$
  Which implies that $\delta_0 < \delta_1 $. With this, we see that as $n\to \infty$, $\delta \to \infty$ and it implies that the function is not continuous at $(x,y)=(0,0)$. 

(It is unclear what this mysterious $\delta$ does here.)

Is my proof correct? (Not yet.) I guess that the only step that seems a little wishy-washy is where I marked with (No. There is a huge space for improvement of writing).
A: [Not a direct answer to your question.]
You are making things too complicated. Being continuous at $(0,0)$ means
$$
\lim_{(x,y)\to(0,0)}f(x,y)=f(0,0)=0.
$$
Now look at the function $f$ on the line $y=x$. 
A: I dk how you inferred $x_0^2+y_0^2<x_1^2+y_1^2$ but it can't be logical.  
For example if $1/n=x_n=ny_n$ then $|f(x_n, y_n)|<1/n$ and $x_n^2+y_n^2>x_{n+1}^2+y_{n+1}^2.$
For another example if $1/n=x_n$ and $0=y_n$. 
To show discontinuity at $(0,0)$ it suffices to find  sequences $((x_n,y_n))_{n\in \mathbb N}$ and $((x'_n,y'_n))_{n\in \mathbb N}$ in $\mathbb R^2 \backslash \{(0,0)\},$ both converging to $(0,0),$ such that the sequence $(\;f(x_n,y_n)-f(x'_n,y'_n)\;)_{n\in \mathbb N}$ does not converge to $0.$
For example $(x_n,y_n)=(0,1/n)$ and $(x'_n,y'_n)=(1/n,1/n).$
A: Comment
The (**) inequality does not hold. For $a=-2,\ b=-1$ we have
$$a<b$$
BUT
$$\frac{a}{b} = 2$$
so it's not true that
$$a<b \implies 1> \frac{a}{b}$$
Possibly you meant positive $a,b$...?
Answer to the problem
For a proof of discontinuity it's enough to demonstrate for arbitrary small neighborhood $N$ of $(0,0)$ there is a point in $N$ at which $f$ differs from $0$ more than by any chosen $\epsilon$. Such (counter)example point is $(t,t)$ for arbitrary $t$:
$$f(t,t) =\frac{2\,tt}{t^2+t^2} = \frac{2t^2}{2t^2} = 1$$
which does not converge to zero with decreasing $|t|$.
Answer to the question
An anwer to your question ('Is my proof correct?') in short is: No, it isn't.
Your attempt doesn't actually prove anything. And as Jack said in the answer there's much more to fix than just the '(?)' deduction.
For example you said 'Let $\epsilon_0=\frac 1n$, then we have $x_0,y_0$ such that $|f(x_0,y_0)|<\frac 1n$.'
However, you just declared 'we have', but you have not shown such $x_0,y_0$ actually exist, which satisfy the inequality.
In fact, there exist both such points at which $f=0$ which makes $|f|<\epsilon_0$ (for example, $(x_0,y_0)=(0,\epsilon_0)$ is such point) and at which $f=1$ which makes $|f|>\epsilon_0$ (the point $(x_0,y_0)=(\epsilon_0/10,\epsilon_0/10)$, for example).
Next you said 'Then we have a $\delta_0$ and $x_0^2+y_0^2<\delta_0^2$'.
But how?! You didn't specify $x_0,y_0$ or at least any bounding for them. You also did not specify what value the $\delta_0$ is, either in relation to (still undefined) $x_0,y_0$ or to $\epsilon_0$.
It is also unclear, what actually your $\delta_0$ is (despite its unknown value). In an epsilon-delta continuity definition the symbol denotes a radius of some neighborhood of the point of limit, and we require the function not to exceed some boundings in that neighborhood. But you do not demonstrate any correlation of your $f$ values' range with respect to $\delta_0$. Even if you defined some concrete values for $x_0,y_0$, it is still a single point and nothing is known about the behavior of $f$ at other points of the delta-neighborhood of $(0,0)$.
The next step of your supposed proof is equally vague, and when you finally come to $\delta_0<\delta_1$ I can only say you are comparing nothing to nothing, and you do that with no sound reason.
Solution
Let me start from the beginning. The definition of continuity, when expressed with words instead of symbols, is:

The real function $f$ defined in some domain $D$ is continuous
  at the point $p$ in $D$ if for any (understood: arbitrarily small)
  positive $\epsilon$ there exists such (small enough) $\delta$,
  that for any point $q$ in $D$, which is not farther than $\delta$ from $p$,
  $f(q)$ differs by no more than $\epsilon$ from $f(p)$.

Even shorter:

If you take a neighborhood of $p$ small enough (radius $\delta$), you get
  values of $f$ in the whole neighborhood arbitrarily close to $f(p)$
  (difference smaller than any chosen $\epsilon$).

Formally we write it as
$$\forall{\epsilon >0}\ \exists{\delta >0}\ \forall q\in D\ \left(\left\Vert q-p\right\Vert<\delta \implies |f(q)-f(p)|<\epsilon\right)$$
Now, if you want to prove continuity at $p$, you need to demonstrate the above condition is satisfied; that is, for any, arbitrarily small chosen bounding $\epsilon$, there is such small distance $\delta$, that within a radius of $\delta$ from $p$, the function does not deviate from $f(p)$ more than by $\epsilon$.
On the other hand, if you want to prove discontinuity, you need to show the opposite; that is, there exists such $\epsilon$ that in each neighborhood of $p$ (which means also within each, arbitrarily small radius $\delta$) there is such point $q$ (at least one) at which $f(q)$ differs from $f(p)$ by more than $\epsilon$.

Look at your function. At any point $q=(x,y)\ne p=(0,0)$ we have
$$f = \frac{2xy}{x^2+y^2}=2\cdot\frac x{\sqrt{x^2+y^2}}\cdot\frac y{\sqrt{x^2+y^2}} \\
=2\frac xr\cdot\frac yr = 2\cos\theta\sin\theta = \sin(2\theta)$$
where $r,\theta$ are polar coordinates of $q$.
As you can see, the value of $f(q)$ depends on the direction from $p$ to $q$ only, but not on their distance. As a result, $f$ reaches ALL values of the sine function in EACH delta-neighborhood of $(0,0)$.
In other words, there exists such $\epsilon$, for example $\epsilon=\frac 12$, that for each, arbitrarily small radius $\delta$ there exists a point $q$ in the delta-neighborhood of $p$, at which $|f(q)-f(p)|\gt\epsilon$. You just need to choose appropriate direction.
An example of such point is $q=(\delta/2,\delta/2)$, whose distance from $p$ is
$$r=\sqrt{(\delta/2-0)^2 + (\delta/2-0)^2} = \sqrt{2\cdot\delta^2/4}=\delta/\sqrt 2 \lt \delta$$
and at which $f(q)=1$ so $|f(q)-f(p)| = 1 \gt \epsilon=1/2$.
That contradicts the requirement of the continuity definition,
hence the function is discontinuous at $(0,0)$,
Q.E.D.
