I want to know the direction where the directional derivative of the funcion $$f(x,y) = \frac{xy^2}{x^2+y^4}, \quad (x,y) \neq (0,0).$$ is maximal.
And we also define, $f(0,0) = 0$.
Now, by a straightforward limit calculation, I determined that $f$ is not continuous at the origin (this taking $y=\sqrt{x}$ and getting 1/2; and $y=2\sqrt{x}$ getting 4/17 since these 2 limits do not coincide, so the limit does not exist, and so $f$ is discontinuous at the origin).
Since the function is not continuous, I can't use the fact that: $$\frac{\partial f}{\partial v} = \nabla f \cdot v.$$
So I calculated the directional derivative via the limit definition: $$\frac{\partial f}{\partial v} (0,0)= \lim_{h \to 0} \frac{f(hv_x,hv_y) - f(0,0)}{h} = \lim_{h \to 0} \frac{hv_x (hv_y)^2}{h \cdot ((hv_x)^2 + (hv_y)^4)} = \lim_{h \to 0} \frac{v_x v_y^2}{v_x^2 + h^2v_y^4}.$$
So, if $v_x = 0$, the directional is equal to $0$, and when it's not zero, it's equal to $\frac{v_y^2}{v_x}$.
Now, this function near 0 can take positive and negative values, so $0$ is not an extreme value. And if I attempt to use the second derivative test, I find that the critical points are the ones of the form $(v_x,0)$, and the Hessian at these values have a determinant equal to $0$, so I can't use the second derivative test. Now, the only thing left is to use Lagrange multipliers under the restriction that $v$ is a unitary vector.
Any ideas?