Understanding the procedure of a simple function being differentiable If $(x,y) \neq (0,0):$
$$f(x,y) = \frac{x^2\cdot y^2}{x^2+y^2}$$
If $(x,y) = (0,0)$:
$$f(x,y)=0$$
I need to show that this function is differentiable.
Here we used linear approximation formula:
$$f(x,y) = f(x_0,y_0) + f_x (x_0,y_0) + f_y (x_0,y_0) + o(x,y)$$
And the upper should imply differentiability if $\lim_{h\rightarrow 0}\frac{o(x,y)}{h} = 0$.
Here we used the point $(x_0,y_0) = (0,0)$ and $h = \sqrt{x^2+y^2}$.
If I understand correctly, we used that point because we want to show that the function is differentiable in that point ?
And where did we get the form for $h$, is it always the same, so $\sqrt{x^2+y^2}$ ?
 A: As your title says, let's start by understanding behind the scenes in a high level:

Definition: Let $f$ be defined in some neighbourhood of $\boldsymbol{a}\in \mathbb{R}^n$. We say "$f$ is differentiable at point $\boldsymbol a$ if there exists a vector $\boldsymbol{A} = (A_1,A_2,...,A_n) \in \mathbb R^n$ such that
$$f(\boldsymbol{a} + \boldsymbol{h}) - f(\boldsymbol a) = \boldsymbol A \cdot \boldsymbol h + o(|\boldsymbol h|), \tag{$\boldsymbol h\to 0$}$$
where $|\boldsymbol h| = \sqrt{\sum_{i =1}^n h_i^2}$

After this definition, almost immediately given the following theorem:

Theorem: If $f$ is differentiable at point $\boldsymbol a$, then each partial derivative of $f$ exists and
$$f(\boldsymbol{a} + \boldsymbol{h}) - f(\boldsymbol a) = \nabla f(\boldsymbol a) \cdot \boldsymbol h + o(|\boldsymbol h|), \tag{$\boldsymbol h\to 0$}$$
where $$\nabla f(\boldsymbol a) = \left\langle \frac{\partial f}{\partial x_k} \mid k = \overline{1,n} \right\rangle$$

Breaking down all these in to $\mathbb R^2$ case, we have the following

Theorem: $f(x,y)$ is differentiable at point $\boldsymbol a = (a_1, a_2)\in \mathbb R^2$ iff $$f(a_1 + h_1, a_2 + h_2) - f(a_1, a_2) = f'_x(a_1, a_2)(x-a_1) + f'_y(a_1, a_2)(y-a_2) + o\left(\sqrt{h_1^2+h_2^2}\right)$$



In your example, with $a = (0, 0)$, you should have
$$f(x,y) - f(0, 0) = f_x'(0,0)x + f_y'(0,0)y + o\left(\sqrt{x^2 + y^2}\right)$$
because $\boldsymbol h = \boldsymbol x - \boldsymbol a$.
Equivalently, if you substitute
$$L(x,y) = f(0,0) + f_x'(0,0)x + f_y'(0,0)y$$
we should have
$$\lim_{(x,y)\to(0,0)}\frac{f(x,y) - L(x,y)}{\sqrt{x^2+y^2}} = 0$$
Now, calculate this limit with your particular function. If it is $0$, you are done!
A: Let me write $\mathbf{v}=(x,y)$. The function is differentiable at $\mathbf{0}$ if and only if there exists a linear map $L\colon\mathbb{R}^2\to\mathbb{R}^2$ such that
$$
\lim_{\mathbf{v}\to0}\frac{f(\mathbf{v})-L(\mathbf{v})}{\|\mathbf{v}\|}=0
$$
because $f(\mathbf{0})=0$.
Can we guess what $L$ should be? Well, the partial derivative with respect to $x$ (outside of $\mathbf{0}$) is
$$
\frac{\partial f}{\partial x}=\frac{2x(x^2+y^2)-2x^3}{(x^2+y^2)^2}y^2=\frac{2xy^4}{(x^2+y^2)^2}
$$
Since $y^2/(x^2+y^2)$ is bounded, we see that the limit at $\mathbf{0}$ is $0$. Similarly for the other partial derivative. So the guess is that $L$ should be the zero map. OK, now we try the limit
$$
\lim_{(x,y)\to(0,0)}\frac{x^2y^2}{(x^2+y^2)^{3/2}}=
\lim_{(x,y)\to(0,0)}|x|\sqrt{\frac{x^2}{x^2+y^2}\frac{y^2}{x^2+y^2}\frac{y^2}{x^2+y^2}}=0
$$
because the big square root is bounded.
