I thought this task up myself, so I'd be good to know whether my solution is correct or not. :-)
Given
$$f(x, y) := {x^3 - y^3 \over x^2 + y^2}$$
for $(x, y) \in \Bbb R \setminus {0},$
$$f(x, y) := 0$$
for $(x, y) = (0, 0),$
I want to examine whether the function is partial differentiable at $(0,0)$ or not.
I need to put in $(x, y) = (0, 0),$ such that:
$\lim_{h\to0}$ (${(x + h)^3 - y^3} \over {(x + h)^2 + y^2}$ $-$ ${x^3 - y^3} \over {x^2 + y^2}$) $\frac 1 h =$ $ \lim_{h\to0}$ (${(0 + h)^3 - 0^3} \over {(0 + h)^2 + 0^2}$ $-$ ${0^3 - 0^3} \over {0^2 + 0^2}$) $\frac 1 h =$
$\lim_{h\to0}$ (${h^3} \over {h^2}$ $-$ $0$) $\frac 1 h =$ $\lim_{h\to0}$ $1 = 1.$
Therefore, the function is differentiable at$ (x, 0).$ For $(0, y),$ the calculation works similar.
Furthermore, I know the algorithm to solve such excercises (at least I think so), but I still have a lack of understanding why it works like this. Why do I have to put in $(0,0)$ for example? And why does this only give me differentiability at $(x, 0)$ first?
Your proof above isn't a proof that $f(x,y)$ is differentiable at $(x,0)$. Your proof is a proof that $f(x,y)$ is partial differentiable at $(0, 0)$. This is the regular definition of derivative: $$\lim_{h \to 0} \ (f(x+h)-f(x))\frac 1 h$$ If I wanted to prove that $f(x)$ is differentiable at $x=0$, I would need to substitute $x=0$ into the limit and then prove the limit exists. You're doing something very similar, except for partial derivatives. In the example you give in your question, your calculating the following limit: $$\lim_{h \to 0} \ ((f(x+h, y)-f(x, y))\frac 1 h$$ Here, only $x$ changes in the arguments while $y$ stays constant, so this is the partial derivative with respect to $x$. Now, you want to prove partial differentiability at $(0, 0)$, so now, you need to substitute $x=0$ and $y=0$ and then prove the limit exists. For partial differentiability with respect to $y$, you do something very similar, except $x$ stays the same and $y$ changes.
Since you never substitute $x=x$ and $y=0$, you are not proving differentiability at $(x, 0)$. Also, since you never substitute $x=0$ and $y=y$, you are not proving differentiability at $(0, y)$. You always do $x=0$ and $y=0$, which means you are proving differentiability at $(0, 0)$. However, in one limit, you are proving partial differentiability with respect to $x$ and in the other, you are proving partial differentiability with respect to $y$.