Sufficient to show the cases when $x = 0$, $y=0$, and $(x,y) \ne (0,0)$? 
For a fixed $k \in \mathbb{N}$, define $f_k: \mathbb{R}^2 \rightarrow \mathbb{R}$ by:
$$
f_k(x,y)=
\begin{cases}
\dfrac{x^2(x+y^2)}{x^2+y^{2k}} &, (x,y)\neq (0,0)\\
0 &, (x,y)=(0,0)\\
\end{cases}
$$
Show that $f_1$ is not differentiable at $(0,0)$, but $f_k$ is differentiable at $(0,0)$ for each $k\geq 2$.
(Hint: At some point it may help to separately consider the cases $|x|\geq |y|^k$ and $|x|\leq |y|^k$.)

So I was able to determine that if $f_k$ is differentiable, the derivative is $(1,0)$. So after some algebra, I see that I have to show that
$\left|\dfrac{x^2y^{2} - x y^{2k}}{(x^2 + y^{2k})\sqrt{x^2+y^2}}\right| \to 0$ as $(x,y) \to (0,0)$.
When evaluating this limit, is it sufficient to show the cases when $x = 0$, $y=0$, and $(x,y) \ne (0,0)$? Why or why not?
 A: So far you have shown that
$$\lim\limits_{(x,y)\to(0,0)}|\frac{f_k(x,y)-f_k(0,0)-\frac{\partial f_k}{\partial x}(0,0)x-\frac{\partial f_k}{\partial y}(0,0)y}{\sqrt{x^2+y^2}}|=\lim\limits_{(x,y)\to(0,0)}|\frac{x^2y^2-xy^{2k}}{(x^2+y^{2k})\sqrt{x^2+y^2}}|$$
The function is differentiable at $(0,0)$ if and only if the above limit is zero.
Note that for the case $k=1$ the function $f_1$ indeed isn't differentiable since along the path $x=y$ we have
$$\lim\limits_{(y,y)\to(0,0)}|\frac{x^2y^2-xy^2}{(x^2+y^2)\sqrt{x^2+y^2}}|=\lim\limits_{(y,y)\to(0,0)}|\frac{y^4-y^3}{2\sqrt2y^3}|=\frac{1}{2\sqrt{2}}\neq0$$
Now suppose that $k\geq2$. Then, using the standard inequality $|\frac{x-y}{2}|\leq\sqrt{\frac{x^2+y^2}{2}}$, we have
$$|\frac{x^2y^2-xy^{2k}}{(x^2+y^{2k})\sqrt{x^2+y^2}}|\leq\sqrt{2}\cdot\sqrt{\frac{x^4y^4+x^2y^{4k}}{(x^2+y^{2k})^2(x^2+y^2)}}\leq\sqrt{2}\cdot\sqrt{\frac{x^2y^4(x^2+y^{2k})}{(x^2+y^{2k})^2(x^2+y^2)}}\leq\sqrt{2}|y|$$
and so
$$\lim\limits_{(x,y)\to(0,0)}|\frac{x^2y^2-xy^{2k}}{(x^2+y^{2k})\sqrt{x^2+y^2}}|\leq\lim\limits_{y\to0}\sqrt{2}|y|=0,$$
i.e. $f_k$ is differentiable at $(0,0)$ for $k\geq2$.
$$$$
