# Determine differentiability of $f(x,y)=\frac{x^3y}{x^4+y^2}, f(0,0)=0$ at the origin

I want to determine the differentiability of the following function at the origin $(0,0)$: $$f(x,y)=\begin{cases} \frac{x^3y}{x^4+y^2} & x \neq 0, y\neq 0 \\ 0, & x=y=0\end{cases}$$ I proved that the partial derivatives of the function, as well as its directional derivatives, at $(0,0)$ all equal zero using the limit definition.

The definition for differentiable that I use is: $$\lim_{h\rightarrow 0}\frac{f(a+h) - f(a)-Ah}{||h||}=0$$ I haven't studied the polar notation yet, so if possible please refrain from using it.

Because of the partial derivatives, $A=(0,0)$ although I couldn't get very far with that: $$\lim_{h\rightarrow 0}\frac{f((0,0)+(h_x,h_y)) - 0-0}{||h||}=\lim_{h\rightarrow 0}\frac{h_x^3\cdot h_y}{(h_x^4+h_y^2)(h_x^2+h_y^2)^{1/2}}$$

And because of the coefficients, I couldn't find a path that would help me disprove it (as is done often,$(t,t),(t-t)(1/n,1/n^2),(t,t^2)$).

So I looked at the two pages of questions regarding this function here on MO. Most of them focused on showing that it is continuous at the origin, which is nice but insufficient.

This says that it is not differentiable, but the path doesn't seem right $$\frac{t^3t^2}{t^4+t^4}=\frac{t^5}{2t^4}=\frac{t}{2}$$ I didn't find it in the book available online.

Looking at the graph however, it is more similar to the non-diffetiable exampe from math-insight but going by the drawing is not enough.

So, can you help me either prove or disprove its differntiablity at the origin? Don't do my homework though, just a little help with the next step or pointing to an error I made would be much appreciated. Thanks.

As you have already said, your function is differentiable if and only if the function $$\varphi(x,y) := \frac{f(x,y)}{\sqrt{x^2+y^2}}$$ goes to $0$ as $(x,y) \to (0,0)$. (I use $(x,y)$ instead of $(h_x, h_y)$ to simplify the notation.)
But you see in a moment that along the path $(t,t^2)$ you have $$\varphi(t, t^2) = \frac{t^5}{2 t^4\sqrt{t^2+t^4}} = \frac{t}{2 |t| \sqrt{1+t^2}}$$ and the limit of this restriction as $t\to 0$ does not exist.