can someone verify this
let $f(x,y) = \frac{x^2 y }{x^4 + y^2}$ forall $(x,y) \neq (0,0)$ and $f(0,0) = (0,0)$ find $\textbf{v}$ for which does directional derivative exists at $\textbf{0}$
$D_{\textbf{v}} f(\textbf{0}) = \lim_{t \rightarrow 0} \frac{f(tv_1, tv_2) - f(0,0)}{t}$
$\hspace{1.4cm} = \lim_{t \rightarrow 0} \frac{v_1^2v_2}{t^2 v_1^4 + v_2^2}$
$D_{\textbf{v}}(\textbf{0}) = \begin{cases} (v_1/v_2)^2 & v_2 \neq 0 \\ \infty & v_2 = 0 \end{cases}$
Your solution is not correct.
If $v_2=0$, then $\frac{v_1^2v_2}{t^2 v_1^4 + v_2^2}=0 \to 0$ as $t \to 0.$
$v_2 \ne 0$, then $\frac{v_1^2v_2}{t^2 v_1^4 + v_2^2} \to \frac{v_1^2v_2}{v_2^2}=\frac{v_1^2}{v_2}$ as $t \to 0.$