Answer
The path limit along $x=0$ and $y=kx$ is equal to $0$.
Work Step by Step
We are given that $f(x,y)=\dfrac{2x^2y}{x^4+y^2}$
a) When $x=0$
Then, we have $\lim\limits_{(x,y) \to (0,0)}\dfrac{2x^2y}{x^4+y^2}=0$
b) Next, we have the second approach along $y=kx$
So,
$\lim\limits_{(x,y) \to (0,0)}\dfrac{2x^2y}{x^4+y^2}=\dfrac{2x^2(kx)}{x^4+(kx)^2}=0$
This implies that the path limit along $x=0$ and $y=kx$ is equal to $0$.
