Prove or disprove each of the follow function has limits $x \to a$ by the definition
$$\lim_{(x, y) \to (0,0)}\frac{x^2y}{x^4 + y^2}$$
Let $y = mx^2$
$$\frac{x^2y}{x^4 + y^2}=\frac{x^4m}{x^4(1+m^2)} = \frac{m}{1+m^2}$$
Hence, $$\lim_{(x, y) \to (0,0)}\frac{x^2y}{x^4+y^2} = \frac{m}{1+m^2}$$
Therefore, since it depends on the value of m, the limit does not exist
Correct?
What you wrote looks correct, but, in a sense, is technically incomplete.
Definition (or characterization) of a limit of function is as follows:
$L$ is a limit of $f(x,y)$ at $(a,b)$ iff for every sequence $(a_n,b_n)$ in the domain of $f$ without $(a,b)$ such that $(a_n,b_n) \to (a,b)$, we have $f(a_n,b_n)\to L$.
So, to complete your proof, choose sequences $(a_n, b_n) = (1/n, 0)$ (this corresponds to $m = 0$) and $(a'_n,b'_n) = (1/n, 1/n^2)$ (this corresponds to $m = 1$). Both converge to $(0,0)$, yet, from what you've shown, $f(a_n,b_n)\to 0$ and $f(a_n',b_n')\to 1/2$. Thus, the limit cannot exist.
There is another possible argument, which comes from change of variables for limits:
Assume $\lim_{(x,y)\to(a,b)}f(x,y)$ exists and $\lim_{t\to c} g(t) = (a,b)$. If $g$ is such that $g(t)\neq (a,b)$ on some punctured ball $B(c,\varepsilon)\setminus\{c\}$, then $\lim_{t\to c}f(g(t)) = \lim_{(x,y)\to(a,b)}f(x,y)$.
In your case, you have $g_m(t) = (t,mt^2)$ and $\lim_{t\to 0}f(g_m(t)) = \frac{m}{1+m^2}.$ Now, choose two different values of $m$, say $0$ and $1$, and from what you have $\lim_{t\to 0}f(g_0(t))\neq \lim_{t\to 0}f(g_1(t))$, so the limits cannot exist.
Yes that correct, indeed we can consider for example
for $x=0,\, y\neq 0 \implies \frac{x^2y}{x^4 + y^2}=0$
for $x=t,\,y=t^2, \,t\to 0 \implies \frac{x^2y}{x^4 + y^2}=\frac{t^4}{t^4 + t^4}\to \frac12$