Checking differentiability at $(0,0)$ Check
$$f(x,y) = 
\begin{cases}
\frac{x^3y^3}{x^2+y^2} & x^2+y^2 \neq 0 \\
0 & x = y = 0
\end{cases}$$
is differentiable at $(0,0)$  or not
.
I have a very vague understanding of differentiability in case of multi-variable calculus . I know that if a function $f(x,y)$ is differentiable at a point $(a,b)$ , then it can be uniquely represented in the form
$$f(a+h,b+k)-f(a,b)=Ah +Bk+h\phi(h,k)+k\psi(h,k)$$ where A and B are dependent on the point and the function that we are choosing , while $\phi(h,k)$  , $\psi(h,k)$ tend to $0$ as $(h,k)$ approaches $0$. This doesn't seem to fit here.
Hints please.
 A: The derivative at $(0,0)$ exists if there is a linear operator $Df(0,0) = L$ from $\mathbb{R}^2$ to $\mathbb{R}$ such that
$$\lim_{(h,k) \to (0,0)}\frac{|f(h,k) - f(0,0) - L \cdot (h,k)|}{\sqrt{h^2 + k^2}}= 0$$
Note that
$$\frac{|f(h,k) - f(0,0)|}{\sqrt{h^2+k^2}} = \frac{|h|^3|k|^3}{(h^2+k^2)^{3/2}} = \frac{1}{2^{3/2}}\underbrace{\left(\frac{2|h||k|}{h^2 + k^2} \right)^{3/2}}_{\leqslant 1}|h|^{3/2}|k|^{3/2} \\\leqslant |h|^{3/2}|k|^{3/2} \underset{(h,k) \to (0,0)}\longrightarrow 0$$
Hence the derivative at $(0,0)$ exists and is the zero operator,
$$Df(0,0): (h,k) \mapsto 0$$
You could have obtained a clue that this was a suitable candidate for the derivative by noting that the partial derivatives are both $0$, but this alone is not sufficient for existence.  A sufficient condition is that the partial derivatives exist in a neighborhood of a point and are continuous at the point.
A: If $(D_uf)(0,0)$ is the directional derivative of $f$ at $(x,y)=(0,0)$ in the direction of the unit vector $u=(u_1,u_2)$ you can compute $(D_uf)(0,0)$ directly as the following limit: $$(D_uf)(0,0)=\lim_{t \rightarrow 0}\frac{f(u_1t,u_2t)-f(0,0)}{t}=\lim_{t\rightarrow 0}u_1^3u_2^3t^3=0$$ Note that $(D_uf)(0,0)=\vec{0}\cdot u$ is a linear function of $u$, so $f'(0,0)$ exists. If the above limit were not a linear function of $u,$ then $f$ wouldn't be differentiable at the origin. Check here. As pointed out in the comments, a standard way to prove that a function is differentiable at $(0,0)$ is to show that $f_x$ and $f_y$ exist and are continuous at $(0,0)$.
