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?

Let $t=\frac yx$. (Note the we can do this only for $x\ne0$. But $x=0$ should not be a problem, since the whole expression is zero in that case.)
Then we have 
$$\frac{xy(x+y)}{x^2-xy+y^2}=\frac{x^3}{x^2} \cdot \frac{t(1+t)}{1-t+t^2}$$
If we can show that the expression
$$g(t)=\left|\frac{t(1+t)}{1-t+t^2}\right|$$
is bounded, then this is sufficient to show that the limit is zero.
For example:


*

*Can you show that for $t\le\frac12$ you have $|t^2-t| \le t^2-t+1$, which means that in this case we have $g(t)<1$?

*Can you show that for $t>\frac12$ the value $g(t)$ is positive and $t^2+t<3(t^2-t+1)$, which means that $g(t)<3$?


Or simply try some other way to show that this expression is bounded.
If will add plot form WA:

