Wrong Wolfram|Alpha limit? $ f(x,y) = \frac {xy}{|x|+|y|} $ for $(x,y)\to(0,0)$

Pretty simply, we have $$ |xy|=\max(|x|,|y|)\min(|x|,|y|)\tag{1} $$ and $$ |x|+|y|\ge2\min(|x|,|y|)\tag{2} $$ Therefore, $$ \left|\frac{xy}{|x|+|y|}\right|\le\frac{\max(|x|,|y|)}{2}\tag{3} $$ Thus, $$ \lim_{(x,y)\to(0,0)}\left|\frac{xy}{|x|+|y|}\right|\le\lim_{(x,y)\to(0,0)}\frac{\max(|x|,|y|)}{2}=0\tag{4} $$

You are right, though you mix up the direction of proof (by what you write, you literally just show "if the limit exists, then it is $0$").

Given $\epsilon>0$, let $\delta=\epsilon$. Assume $(x,y)\ne(0,0)$ is a point with $|(x,y)|<\delta$. Then especially $0<r<\delta$ with $r:=\max\{|x|,|y|\}$ and hence $$ \left|\frac{xy}{|x|+|y|}\right|=\frac{|x|\cdot|y|}{|x|+|y|}\le \frac{r^2}{r+0}=r<\delta<\epsilon,$$ as was to be shown, i.e. $$ \lim_{(x,y)\to(0,0)}\frac{xy}{|x|+|y|}=0.$$

You are right, wolfram is wrong. It might happen...

Only you should correct your exposition of the definition. You say:

By definition, if blah blah, then bleh bleh

you should say:

By definition, blah blah, if bleh bleh

In fact you prove bleh bleh to have blah blah.

The first thing to do when computing this kind of limits is trying to isolate a bounded expression.

Assuming $(x,y)\ne(0,0)$ in what follows, we clearly have

$$ \left|\frac{y}{|x|+|y|}\right|\le 1. $$

Therefore we can write

$$ -|x|\le\frac{xy}{|x|+|y|}\le |x| $$

and so

$$\lim_{(x,y)\to(0,0)}\frac{xy}{|x|+|y|}=0$$

follows by the squeezing theorem.

Hint: By the arithmetic-geometric inequality $$ \frac{|xy|}{|x|+|y|}\leq\frac{\sqrt{|xy|}}{2}. $$