Does the limit $\lim_{(x,y)\to (0,0)} \frac{x^3y^2}{x^6+y^4}$ exists? I was trying to find if the the following limit exists:
$$
\lim_{(x,y)\to (0,0)} \frac{x^3y^2}{x^6+y^4}
$$
From this topic I tried to switch to polar coordinates:
$$
\frac{x^3y^2}{x^6+y^4}=\frac{r^3\cos^3\theta\cdot r^2\sin^2\theta}{r^6\cos^6\theta+r^4\sin^4\theta}=\frac{r\cos^3\theta\sin^2\theta}{r^2\cos^6\theta+\sin^4\theta}\to 0
$$
But apparently this limit does not exists (proved it in another way). Why this method worked for the limit in that topic and here it didn't work?

Consider the path along the x-axis, y-axis and $y=x^{\frac{3}{2}}$.
Along the x-axis:
$$\lim_{(x,0) \to (0,0)} \frac{x^3 \cdot 0}{x^6 +  0}=0
$$
Along the y-axis:$$\lim_{(0,x) \to (0,0)} \frac{0 \cdot y^2}{0 +  y^4}=0
$$
Along the curve $y=x^{\frac{3}{2}}$:
$$ \lim_{(x,x^{\frac{3}{2}}) \to (0,0)} \frac{x^3{(x^{\frac{3}{2}})}^2}{x^6 + {(x^{\frac{3}{2}})}^4 }=\frac{x^6}{2x^6}=\frac{1}{2}$$
Therefore, the limit does not exist.
