Prove $f(x,y) = xy/(x^2 + y^2)$ is continuous everywhere except $(0,0).$ I'd just like to ask you if my proof here is valid. I'll provide you with the method I used and if it seems ok let me know! If not, explanations would be helpful!
My main approach to this question involves using the theorem that if functions $g$ and $h$ are both continuous, then their product and their sum are also continuous.
I begun by splitting $f$ into $(g)(1/h)$, where $g(x,y) = xy$ and $h(x,y) = x^2 + y^2.$
Then, I again split $g$ into $2$ functions $p(x) = x$ and $q(x) = y$ and claimed that since they were both polynomials of order 1, they must be continuous everywhere. (Is it ok to just say this? It's a theorem in my book.) Hence, their product $xy$ must be continuous.
Similarly, the sum of $x^2$ and $y^2$ (two polynomials of order $2$) must also be continuous.
Now, we are left with $f(x,y) =$ the product of $g$ and the inverse of $h$. The inverse of $h$ is again continuous everywhere, except $(0,0)$ as $h(0,0)$ is undefined.
Thus, $f(x,y)$ is continuous everywhere, except $(0,0)$.
Thanks for reading!
 A: In a simplified and quicker approach, just consider those points where $f$ is not well defined, to identify non-continuity. You need more care in your discussion on " $h$ is not defined". In this case, the only point which gives "trouble" for $f$ is $(0,0)$. But we must check this formally! Using polar coordinates $(\rho,\theta)$ you can prove that
$$\lim_{(x,y)\rightarrow (0,0)}f(x,y) ~~(*)$$
is equal to the limit
$$\lim_{\rho\rightarrow 0}\cos\theta\sin\theta:=q(\theta), $$
which is a function of the angle $\theta$: in summary the limit $(*)$ depends on the path chosen to approach $(0,0)$, which means that $f$ is not continuous at $(0,0)$.
For all points different from $(0,0)$ just use the definition of continuity of a function $f:\mathbb R^2\rightarrow \mathbb R$ (with $\epsilon-\delta$, for example); the thesis follows.
A: Disclaimer: I don't agree on the conclusion that $f$ is not continuous at the origin unless you define $f(0,0)$ in some way.
This said, I can't really understand your question. Is it

Prove that $f$ is continuous at every point of $\mathbb{R}^2 \setminus \{(0,0\}$

or

Provethat the only discontinuity point of $f$ is $(0,0)$?

In the first case, your solution is pretty good, since you picked a point $(x,y) \in \mathbb{R}^2 \setminus \{(0,0\}$ and proved that $f$ is continuous at this point.
In the second case, your proof lacks the detail that $f$ is discontinuous at the origin. You can show this by approaching $(0,0)$ along the straight line $y=mx$, so that $f(x,mx)=\frac{mx^2}{(1+m^2)x^2}=\frac{m^2}{1+m^2}$, which depends on $m$. Hence the limit $\lim_{(x,y) \to (0,0} f(x,y)$ cannot exist.
