Show limit of function $\frac{xy(x+y)}{x^2-xy+y^2}$ for $(x,y)\to(0,0)$ Show $$\lim_{(x,y)\to (0,0)} xy \frac{(x+y)}{x^2-xy+y^2}=0$$
If I approach from $y=\pm x$, I get $0$. Is that sufficient?

No, that is not sufficient. That can only show that if the limit exists, then it should equal $0$. But it doesn't show that the limit exists.
Change to polar coordinate system, so that 
$$
\lim_{(x,y)\to (0,0)} \frac{xy(x+y)}{x^2-xy+y^2} = \lim_{r\to 0} \frac{ (r^2 \sin \theta \cos \theta) (r\sin \theta + r\cos \theta)}{r^2-r^2\sin \theta \cos \theta}= \lim_{r\to 0} \frac{r\sin \theta \cos \theta(\sin \theta + \cos \theta)}{1-\sin \theta \cos \theta}
$$
(Check that two limits are equilvalent from the ($\varepsilon$-$\delta$) definition of the limits)
But notice that $|1-\sin \theta \cos \theta|=|1-\sin (2\theta) /2|\ge 1/2$.
Hence $\displaystyle \frac{\sin \theta \cos \theta(\sin \theta + \cos \theta)}{1-\sin \theta \cos \theta}$ is bounded for all $\theta\in \mathbb{R}$, and so 
$$
\lim_{r\to 0} \frac{r\sin \theta \cos \theta(\sin \theta + \cos \theta)}{1-\sin \theta \cos \theta}=0
$$
