Prove that $\frac{xy}{x^2+y^2}$ is not continuous 
Let $f:\mathbb R\times \mathbb R\to \mathbb R$ a function with two variables, prove that the function is not continuous 
  $$f(x,y)=\begin{cases}\dfrac{xy}{x^2+y^2}, & \forall(x,y)\neq (0,0)\\
\\
0, & (x,y)=(0,0)\end{cases}$$

Note: this question from topology course, not from calculus
Attempt:
Let $X=\mathbb R\times \mathbb R,\\
 Y=\mathbb R\\
(x_0,y_0)=(0,0)\\
f\left(x_0,y_0\right)\in U\subseteq Y$
Now I should show that $\exists V\subseteq X$ such that $x\in V$ and $f(V)\subseteq U$
How can I prove that?
 A: Hint. What happens along the lines $y=0$ and $y=x$?
A: If you let $x=0$ and take the limit as $y \to 0$, you get $0/y^2= 0$.
If you let $x=y$ and take the limit as $y \to 0$ you get $1/2$.
A: Consider $(x,y)\to(0,0)$ in various radial directions, say along $y=kx$. 
Note that the limit will then depend on the direction of approach (i.e., on $k$) so that the limit does not exist. To exist, it must be independent of the manner of approach.
A: If you're interested in proving the non-continuity of a function at a certain point, try to find two different paths on which to approach the point in question (i.e. take two different $\lim$) along the function graph which give two different values. Remember the definition of continuity. Hope this helps 
A: To give a direct topological proof that the function is not continuous at the origin, suppose it were continuous and derive a contradiction.  So, using the continuity assumption with $U = B_{1/3}(0)$, you would be able to find a neighborhood $V$ of $(0, 0)$ such that $V \subseteq f^{-1}(U)$.  Since $V$ is a neighborhood, then for some $\delta > 0$, we have $B_\delta((0, 0)) \subseteq V$.  Then we can conclude that whenever $\lVert (x, y) \rVert < \delta$, $|f(x, y)| < \frac{1}{3}$.  However, note that for any $t \ne 0$, $f(t, t) = \frac{1}{2}$, and it is possible to choose $t$ such that $\lVert (t, t) \rVert < \delta$, which gives the desired contradiction.
(Or, for a slightly less direct proof, you can use the fact that if $f$ were continuous, then we would have that for any sequence $(a_n)_{n=1}^\infty \subseteq \mathbb{R}^2$, if $a_n \to (0, 0)$, then $f(a_n) \to f(0, 0) = 0$.  However, then you can get a contradiction using the sequence $a_n = (\frac{1}{n}, \frac{1}{n})$.)
A: If $f$ were continuous then the pre-image of every open set would be an open set, but also the pre-image of every closed set would be a closed set.
We can see that 
$$
f(x,y)=\frac12 \frac{2xy}{x^2+y^2}=\frac12\implies x^2+y^2=2xy\implies(x-y)^2=0\implies x=y
$$
So the function assumes the value $1/2$ on all the points of the line $x=y$, and in no other points, but excluded $(0,0)$ (where it assumes the value $0$). So 
$$
f^{-1}(1/2)=\{(x,y)\in\mathbb{R}^2:x=y\text{ and }(x,y)\neq(0,0) \}
$$
and $\{1/2\}$ is closed but $f^{-1}(1/2)$ is not.
