calculus - Why is this function not differentiable at $(0,0)$

I was brushing up on some calculus and I was thinking about the following function:

Let $$f(x) = \begin{cases} \frac{yx^{6} +y^{3}+x^{3}y}{x^{6}+y^{2}} & \text{for $(x,y)$ $\neq (0,0)$,} \\ 0 & \text{for $(x,y)$ $=$ $(0,0)$. } \\ \end{cases}$$ The function $f$ is continuous over the entire real line and is differentiable everywhere except at $x=0$.

For some arbitrary nonzero $\xi=(u,v)$, I was able to compute the directional derivative ${\xi}f(0,0)=\frac{1}{h}f(hu,hv)=v$. In particular for $\xi=\frac{\partial}{\partial{x}}=(1,0)$, we have $\frac{\partial{f}}{\partial{x}}(0,0)=0$ and for $\xi=\frac{\partial}{\partial{y}}=(0,1)$, we have that $\frac{\partial{f}}{\partial{y}}(0,0)=1$. From here we see that partial derivative depends linearly on the vector we differentiate on meaning $(t{\xi}))f(0,0)=(tu,tv)f(0,0)=tv=t({\xi}f(0,0))$.

This function is not differentiable at the origin, in fact it is not even continuous at the origin. I was having trouble seeing exactly why. I've forgotten lots of calculus, but I was wondering if the fact that $\frac{\partial{f}}{\partial{y}}(0,0)=1$ while $\frac{\partial{f}}{\partial{x}}(0,0)=0$ implies discontinuity at $(0,0)$.

Also as a side note, there is the big theorem in differential calculus which says if all the directional derivatives ${\xi}f: \mathbb{R}^{2} \rightarrow \mathbb{R}$ are continuous at a point $p$, then $f$ is differentiable at the point $p$. I'm confused as to how the partial derivatives above are not continuous.

Any help is much appreciated.

Answer

Hint: consider approaching the origin along the path $(x,x^3)$