Two-dimensional limit, is my approach correct? The limit is
$$\lim_{(x,y)\to(0,0)}\frac{x^3y}{x^4+y^2}$$
As usual, I tried checking along particular paths, namely the axes and the curves $y=mx^n$ for various values of $n$, but to no avail; all the limits evaluate to $0$. I resorted to converting to polar coordinates, giving
$$\lim_{r\to0^+}\frac{r^4\cos^3t\sin t}{r^2\left(r^2\cos^4t+\sin^2t\right)}$$
In an attempt to apply the squeeze theorem, I found that trig expression in the numerator is bounded, with $|\cos^3t\sin t|\le\dfrac{3\sqrt3}{16}$, and that in the denominator I have
$$\begin{align*}
|r^2\cos^4t+\sin^2t|&\le r^2|\cos^4t|+|\sin^2t|\\
&\le r^2+1
\end{align*}$$
and so
$$\lim_{r\to0^+}\frac{r^4\cos^3t\sin t}{r^2\left(r^2\cos^4t+\sin^2t\right)}=\frac{3\sqrt3}{16}\lim_{r\to0^+}\frac{r^4}{r^4+r^2}=0$$
Is my reasoning valid? I am unsure about whether or not the bound for the denominator is correct.

$$  (x^2 - y)^2 \geq 0,   $$
$$  x^4 - 2 x^2 y + y^2 \geq 0,  $$
$$ x^4 + y^2 \geq 2 x^2 y. $$
$$  (x^2 + y)^2 \geq 0,   $$
$$  x^4 + 2 x^2 y + y^2 \geq 0,  $$
$$ x^4 + y^2 \geq -2 x^2 y. $$
$$ x^4 + y^2 \geq 2 x^2 |y|.  $$
$$ \color{magenta}{ \left| \frac{x^2 y}{x^4 + y^2} \right| \leq \frac{1}{2} }  $$
Your numerator is $x^3 y$ rather than $x^2 y;$ what happens now?
Some detail, not directly needed for your question but probably helpful anyway, at recent Continuity in $\mathbb R^n$
