Answer
The path limit along $x=0$ and $y=kx$ is equal to $0$.
Work Step by Step
Consider $f(x,y)=\dfrac{2x^2y}{x^4+y^2}$
Let us consider the approach: $(x,y) \to (0,0)$ when $x=0$
Then, we get $\lim\limits_{(x,y) \to (0,0)}\dfrac{2x^2y}{x^4+y^2}=\dfrac{2(0)^2y}{(0)^4+y^2}=0$
Let us consider the next approach: $(x,y) \to (0,0)$ along $y=kx$
Then, we get $\lim\limits_{(x,y) \to (0,0)}\dfrac{2x^2y}{x^4+y^2}=\dfrac{2x^2(kx)}{x^4+(kx)^2}=0$
This shows that the path limit along $x=0$ and $y=kx$ is equal to $0$.
