# What is the correct method to show that $f(x,y)$ is not-differentiable at $(0,0)$?

I'm trying to prove that $f$, defined as $$f(x,y)=\left\{\begin{array}{cl} \dfrac{x^2y}{x^2 + y^2}&(x,y)\neq (0,0) \\ 0&(x,y)=(0,0)\end{array}\right. ,$$ is not differentiable at $(0,0)$. Is the correct way to approach such problems just to show that $$\lim_{(h,k)\to(0,0)}\frac{f(h,k)-f(0,0)-\mathbf{A}(h,k)}{\sqrt{h^2+k^2}}\neq​0\ ?$$‌ Or should I show that that limit does not exist?

I was told that even if the limit turns out to be non-zero, even then the function will be differentiable at $(0,0)$. Is that true? I mean if the limit doesn't turn out to be $0$ doesn't it mean that the tangent plane is not the best linear approximation for the function? If tangent plane isn't the best linear approximation, how can it be differentiable at that point (given the fact that the tangent plane is defined as the derivative)?

Also, should I take $A$ as just the jacobian $[\frac{\partial f}{\partial x} \space \frac{\partial f}{\partial y}]$ and $A(h,k)$ as $[\frac{\partial f}{\partial x} \space \frac{\partial f}{\partial y}][h \space k]^T$ ?

By definition, the function $f$ is differentiable at $(0,0)$ if there exists a vector $\mathbf{A} = (p,q)$ such that $$\lim_{(h,k)\to (0,0)} \frac{f(h, k) - f(0,0) - p h - q k }{\sqrt{h^2+k^2}} = 0.$$ In this case, one also have that $f$ admits partial derivatives at $(0,0)$ and that $\frac{\partial f}{\partial x}(0,0) = p$, $\frac{\partial f}{\partial y}(0,0) = q$.
Since your function is partially differentiable at $(0,0)$ with both partial derivatives equal to $0$, you have that it is differentiable at $(0,0)$ if and only if $$\lim_{(h,k)\to (0,0)} \frac{f(h, k) - f(0,0)}{\sqrt{h^2+k^2}} = 0,$$ i.e., if and only if $$\lim_{(h,k)\to (0,0)} \frac{h^2 k}{(h^2+k^2)^{3/2}} = 0.$$ But you see in a moment that the above limit does not exists. Indeed, if you evaluate the argument of the limit at $(h,h)$, for $h\neq 0$, you get $h^3 / |h|^3$, and this function does not admit limit for $h\to 0$.