How can I show that the limit of this function under these conditions does not exist? Show that the limit of the function, $f(x,y)=\frac{xy^2}{x^2+y^4}$, does not exist when $(x,y) \to (0,0)$.
I had attempted to prove this by approaching $(0, 0)$ from $y = mx$, assuming $m = -1$ and $m = 1$. The result was $f(y, -y) = \frac{y}{1+y^2}$ and $f(y, y) = \frac{y}{1+y^2}$ as the limits which are obviously different. Essentially, I was just wondering what is the correct working out for a solution to this question.
 A: I assume that you would like to see either that the function is not continuous at $(0,0)$ or that the function is not defined at $(0,0)$.
That the function is not defined at $(0,0)$ holds because we do not have division by zero in the real field. 
Let $f: (x,y) \mapsto xy^{2}/(x^{2} + y^{4})$ on $\mathbb{R}^{2}\setminus \{ (0,0) \}$ and let $f(0,0) := 0$.
We claim that the function is not continuous at $(0,0)$. Note that along every straight line through the origin it holds that $f(x,y) \to 0$ as $(x,y) \to (0,0)$, i.e. for every $m \in \mathbb{R}$, along  set $S_{m} := \{ (x,y) \mid y = mx \}$ we have $f(x,y) \to 0$ as $(x,y) \to (0,0)$, i.e. $f\mid_{S}(x,y) \to 0$ as $(x,y) \to (0,0)$. Thus $f(0,0)$ is the limit of $f$ at $(0,0)$ along every straight line through the origin. However, since the function $f$ restricted on some curve, say on the parabola $T := \{(x,y)\mid y^{2} = x \}$, takes the value 1/2, and since for every neighborhood $U$ of $(0,0)$ we have $U \cap T \neq \varnothing$, it follows that $f\mid_{T}(x,y) \to 1/2 \neq f(0,0)$ as $(x,y) \to (0,0)$, whence $f$ is not continuous at $(0,0)$.
I hope the above argument is helpful.      
Please note that if your original question is to ask if the limit of $f$ at $(0,0)$ exists then the above argument also covers your question, for it shows that the limit of $f$ at $(0,0)$ may or may not exist, dependent on the choice of paths along which we are talking about the limit. 
A: $$\lim\limits_{(x,y)\to (0,0)}\frac{xy^2}{x^2+y^4}$$
Using polar coordinates, we have
$$\lim\limits_{r\to 0^+}\frac{r^3\cos\phi\sin^2\phi}{r^2\cos^2\phi+r^4\sin^4\phi}$$
$$=\lim\limits_{r\to 0^+}\frac{r\cos\phi\sin^2\phi}{\cos^2\phi+r^2\sin^4\phi}$$
Now lets try to find bounds that are independent of $\phi$. Since 
$$\left|\cos\phi\sin^2\phi\right|\leq 1$$
We have
$$\frac{r\left|\cos\phi\sin^2\phi\right|}{\left|\cos^2\phi+r^2\sin^4\phi\right|}\leq \frac{r}{\left|\cos^2\phi+r^2\sin^4\phi\right|}$$
Notice that the right hand side cannot be bounded in terms that are independent of $\phi$. Therefore
$$\lim\limits_{(x,y)\to (0,0)}\frac{xy^2}{x^2+y^4}=\mbox{non existent}$$
