<s>[INST] Prove that $f(x,y)$ is continuous in $(0,0)$ Prove that $f(x,y)$ is continuous in $(0,0)$, where
\begin{equation}
f(x,y) = \begin{cases}
\frac{x^2y}{x^4+y^2}, & (x,y)\neq 0\\
0, & (x,y) = (0,0)
\end{cases}
\end{equation}
The solution I have is that f is not continuous in $(0,0)$. (The solution doesn't say more than that.)
However, the result I got is that $f$ is continuous in $(0,0)$. Here's my approach:
Lets transform $x$ and $y$ into their polar coordinates, so that we can approach $(0,0)$ from any direction by varying $\theta$: 
$(x,y) = (r\cos(\theta),r\sin(\theta)) \qquad r\in\mathbb{R}^+_0 \quad \theta\in[0,2\pi)$
Then $f$ is continuous iff
\begin{equation}
\lim_{(x,y)\to (0,0)} f(x,y) = 0 = f(0,0)
\end{equation}
By using the polar coordinates and letting $r\to 0$ we get:
\begin{equation}
\lim_{(x,y)\to (0,0)} f(x,y)
=
\lim_{r\to 0} \frac{r^3\cos^2\theta\sin\theta}{r^4\cos^4\theta + r^2\sin^2\theta}
= \lim_{r\to 0} \frac{r\cos^2\theta\sin\theta}{r^2\cos^4\theta + \sin^2\theta}
\end{equation}
By a case distinction by $\theta$ we get:


*

*$\theta\in[0,2\pi)\backslash\{0,\pi\}$:
\begin{equation}
\lim_{r\to 0} \frac{r\cos^2\theta\sin\theta}{r^2\cos^4\theta + \sin^2\theta}
=
\lim_{r\to 0} \frac{0}{\sin^2\theta} = 0 = f(0,0)
\end{equation}

*$\theta\in\{0,\pi\}$: $\Longrightarrow \sin\theta = 0$
\begin{equation}
\lim_{r\to 0} \frac{r\cos^2\theta\sin\theta}{r^2\cos^4\theta + \sin^2\theta}
=
\lim_{r\to 0} \frac{0}{r^2\cos^4\theta} = 0 = f(0,0)
\end{equation}


Form 1. and 2. we can conclude that $f$ is continuous in $(0,0)$.
What am I doing wrong in my approach?
 [/INST] You have shown the limit is $0$ along all straight lines through $(0,0).$ That's not enough: There are tons of paths to $(0,0)$ that are not straight lines. For example, the path $(x,x^2)$ as $x\to 0.$ 
</s>