Check if the function is differentiable or not 
Consider $f: \mathbb{R^2} \to \mathbb{R}$, $$f(x, y) = \begin{cases}\frac{xy^2}{x^2 + 2y^2} & (x, y) \ne (0,0)\\0 & (x, y) = (0,0)\end{cases}$$
Where is $f(x, y)$ differentiable over its domain?

I am considering $(0, 0)$ as a point, but I am not sure how to go about proving it (or disproving)
 A: Use the definition of partial derivatives to compute
$$\frac{\partial f}{\partial x}(0,0)=\lim_{x\to 0}\frac{f(x,0)-f(0,0)}{x-0}=0$$ and $$\frac{\partial f}{\partial y}(0,0)=\lim_{y\to 0}\frac{f(0,y)-f(0,0)}{y-0}=0.$$
Then to check differentiability
\begin{align}\lim_{(x,y)\to (0,0)}\frac{f(x,y)-\frac{\partial f}{\partial x}(0,0)(x-0)-\frac{\partial f}{\partial y}(0,0)(y-0)}{\sqrt{(x-0)^2+(y-0)^2}}\\
=\lim_{(x,y)\to (0,0)}\frac{xy^2}{(x^2+2y^2)\sqrt{x^2+y^2}}.
\end{align}
To prove differentiability you need to check if the limit exists and is zero. In this case the limit does not exist. Hint: use restrictions, such that $y=x$ or $y=0$,
A: The partial derivatives at $(0,0)$ are both $0$. Hence, $f$ is differentiable at $(0,0)$ if and only if the following limit exits:
$$\lim_{(h,k)\to(0,0)} \frac{f(h,k)}{\sqrt{h^2+k^2}}$$
This limit obviously doesn't exist.
A: There are directional derivatives in every direction. These are found immediately in polar when both numerator and denominator are homogeneous, using
$$ r \frac{\cos \theta \sin^2 \theta}{ \cos^2 \theta + 2 \sin^2 \theta}  $$
so that the directional derivative in (unit) direction $\theta$ is
$$  \frac{\cos \theta \sin^2 \theta}{ \cos^2 \theta + 2 \sin^2 \theta}  $$
Now, writing this back in cartesian coordinates, does this indicate a linear function of the direction (not necessarily unit)? The answer is no, as this is zero on the axes, when $\theta $ is a multiple of $\pi/2,$ but nonzero elsewhere. If the thing were linear, we would always get zero.
