$\lim_{(x,y) \to (0,0)} {\frac{x^2y}{x+xy+y^2}}$ I was playing around with limits of two real variables, when I came up with the one on the title.
I tried all directional limits and they were all $0.$
So then I tried a few relative limits like $y=ax^2$ , $y=x^3-x^2$ and $0\leq x=\sqrt{y}$, but still got $0.$
 I tried then with $y=\cos(x)-1,$ and it gave a complicated expression which I plugged into symbolab, and it gave $0$ too. 
So I tried to prove that it exists and is $0.$ So I'd need to find an inequality between the expression below and a norm $ \vert \vert (x,y) \vert \vert.$ I don't see how to manipulate it however, since the things in the denominator can be negative, and the only inequality I know can be applied here is $|xy| \leq x^2+y^2$, which would get:
$$\left|\frac{x^2y}{x+xy+y^2}\right| \leq \left| \frac{x^2+x^2y^2}{x+xy+y^2} \right|.$$
I don't know how to proceed from that, nor have any other ideas to try to disprove it exists.
If possible give just a hint please, and I haven't yet learned polar coordinates nor partial derivatives.
 A: The key issue here is to find a meaningful lower bound on $|x + xy + y^2|$; this can be written as
$$|x + xy + y^2| = |xy| \left|\frac{1}{y} + 1 + \frac y x\right|.$$
This is problematic if
$$\frac 1 y + \frac y x \approx -1$$
So try thinking about a case where $y \to 0^+$ (in which case the first term blows up), but where $x \to 0^-$ is carefully balanced with $y$ to cause a problem.


 This is the case on a hyperbola and is quite apparent from a plot of the surface.

A: For any open disc $D$ containing $(0,0)$  we can find $(x,y)\in D$ with $x\ne 0\ne y$ and $x+xy+y^2=0....$ I.e. if $y\ne 0$ and $ |y|$ is small enough and $  x=-y^2/(1+y).$
Then for any $n\in \Bbb Z^+$ we can also find $(x_n,y_n)\in D,$ close enough to $(x,y)$ that  $|x^2y-x_n^2y_n|<|x^2y|/2$ (which implies $|x_n^2y_n|>|x^2y|/2$)..... and also such that $0<|x_n+x_ny_n+y_n^2|<1/n .$ 
That is so, because the functions $a(u,v)=u^2v$ and $b(u,v)=u+uv+v^2$ are continuous, and $b(u,v)$ is not constantly $0$ on any non-empty open disc.
Let $g(u,v)=\frac {u^2v}{u+uv+v^2}$ when $u+uv+v^2\ne 0.$ So $|g( x_n,y_n)|>\frac {|x^2y|/2}{1/n}=n|x^2y|/2.$ 
So $\{|g(u,v)|:(u,v)\in D \land u+uv+v^2\ne 0\}\supset \{|g(x_n,y_n)|:n\in \Bbb Z^+\},$ which is a set with no upper bound.
So even when we exclude all $(x,y)$ such that $x+xy+y^2=0,$ the limit in question still does not exist.
Footnote: Regarding the first line above, if $r>0$ and $D=\{(u,v):u^2+v^2<r^2\},$ let $0<y<\min (1/2, r/2)$ and $x=-y^2/(1+y).$ Then $|x|<y^2$ so $x^2<y^4<y^2/4 $ so $x^2+y^2<y^2/4+y^2=5y^2/4<r^2.$
A: This is not a complet answer, but is valid if $x,y>0.$
From $$0<x^2y<x^2y+x^2y^2+xy^3$$ we deduce
$$0<{\frac{x^2y}{x+xy+y^2}}<{\frac{x^2y+x^2y^2+xy^3}{x+xy+y^2}}=xy,$$and with the use of squeeze theorem $$\lim_{(x,y) \to (0^{+},0^{+})} {\frac{x^2y}{x+xy+y^2}}=0.$$
