<s>[INST] Does $\lim_{(x,y) \to (0,0)} \frac{x y \sin^2 y}{x^2 y}$ exists? 
$\displaystyle\lim_{(x,y) \to (0,0)} \frac{x y \sin^2 y}{x^2 y}$

Along $x =y$, it becomes
\begin{equation}
\lim_{(x,y) \to (0,0)}\frac{y^2 \sin^2 y}{y^3} = 0.
\end{equation}
But along $x = y^3$,
\begin{equation}
\lim_{(x,y) \to (0,0)}\frac{y^4 \sin^2 y}{y^7} = \lim_{(x,y) \to (0,0)}\frac{ \sin^2 y}{y^3}.
\end{equation}
The limit is undefined.
This is what I thought, but using wolframalpha, it say the limit is $0$.
Where did I go wrong?
 [/INST] You are right indeed 
$$\frac{x y \sin^2 y}{x^2 y}=\frac{\sin^2y}{y^2 }\frac{y}{x}$$
and


*

*$\frac{\sin^2y}{y^2 }\to 1$
but


*

*$\frac{y}{x}$
has not limit.
</s>