Can I please receive feedback on my proof for the following? Thanks!
$\def\R{{\mathbb R}} \def\x{{\bf x}} \def\f{{\bf f}} \def\0{{\bf 0}}$ Let $f\colon \R^2\to \R$ be given by $$f(\x)=f(x_1,x_2) = \left\{\begin{array}{cl} \frac{x^2_1 x_2}{x^4_1+x^2_2} & \mbox{if $\x\ne\0$,} \\ 0 & \mbox{if $\x=\0$.} \end{array}\right.$$ Show $\displaystyle{\lim_{\x\to\0} f(\x)}$ does not exist.
$\textbf{Solution:}$ Let us consider that $x_1^2 = x_2$. This can be possible even when $\x\to 0$. Therefore, $$\frac{x^2_1 x_2}{x^4_1+x^2_2} = \frac{1}{\frac{x_1^4}{x_1^2x_2}+\frac{x_2^2}{x_1^2x_2}}$$ $$=\frac{1}{\frac{x_1^2 }{x_2}+\frac{x_2}{x_1^2}}$$ $$=\frac{1}{2}.$$
Next, $$\lim_{\x\to\0} f(\x) = \lim_{(x_1,x_2)\to (0,0)} f(x_1,x_2) \text{ as $x_2$ = 0 then}$$ $$=\lim_{(x_1,x_2)\to(0,0)} \frac{x_1^2\cdot 0}{x_1^4+0^2} = \lim_{(x_1,x_2)\to(0,0)} 0 = 0.$$
Observe, we have different limits as $x\to 0$ in different directions. So, $\lim_{\x\to\0} f(\x)$ does not exist.
Choose a curve $x_2=mx_1^2, m\ne 0$ so that the limit equals $\frac{m}{1+m^2}$ which depends on $m$.