The existence of a partial derivative at a point Let $f(x,y) = \frac{xy}{x^2+y^2}$ if $(x,y)\neq(0,0)$ and $f(x,y) = 0$ if $(x,y)=(0,0)$.
By definition, we have $f_x(0,0) = \lim_{h\rightarrow0} \frac{f(h,0)-f(0,0)}{h}=0$.
But we also have that $f_x(x,y) = \frac{y(y^2-x^2)}{(x^2+y^2)^2}$, and from this equation it seems that $f_x(x,y)$ is not defined at $(0,0)$. Why is this the case? I can see that $f_x(x,y) = \frac{y(y^2-x^2)}{(x^2+y^2)^2}$ doesn't take into account that "$f(x,y) = 0$ if $(x,y)=(0,0)$", whereas by the definition, $f_x(0,0)$ is defined $\iff$ $f(0,0)$ is defined. But I still am not sure to fully understand why we cannot use the equation of $f_x$ to determine whether $f_x$ is defined at $(0,0)$ or not.
 A: $f_x$ is defined at $(0,0)$ and has the value $0$. In all other points we have $f_x(x,y) = \frac{y(y^2-x^2)}{(x^2+y^2)^2}$. The expression $\bar f_x(x,y)  = \frac{y(y^2-x^2)}{(x^2+y^2)^2}$ on the RHS of the equation is indeed not defined at $(0,0)$, but this has nothing to do with the existence of $f_x(0,0)$. You have the same situation with $f$: The expression $\bar f(x,y) = \frac{xy}{x^2+y^2}$ is not defined at $(0,0)$, but nevertheless $f$ is defined at $(0,0)$.
That $\bar f_x(x,y)$ is undefined at $(0,0)$ indicates that $(0,0)$ may have a special role. A natural question is whether $\lim_{(x,y) \to (0,0)} \bar f_x(x,y)$ exists. It does not since $\bar f_x(0,y) = 1/y$ is unbouded.
Thus the special role of $(0,0)$ is that $f_x$ is not continuous at $(0,0)$. More precisely, $f_x$ exists in all points $(x,y)$ and is continuous in all points $(x,y) \ne (0,0)$, but not continuous  at $(0,0)$.
A: The partial derivative $f'_x$ is defined everywhere... It is just not possible to use the differentiation rules at $(0,0)$ because in the limit
$$
f'_x(0,0) = \lim_{h\to 0}\dfrac{f(h,0)-f(0,0)}{h},
$$
the two occurrences of $f$ are not computed in the same branch of $f$. This does not mean that the partial derivative is not defined everywhere, in fact
$$
f'_x(x,y)=\begin{cases}
\dfrac{y(y^2-x^2)}{(x^2+y^2)^2} &, (x,y)\ne(0,0)\\
0 &, (x,y)=(0,0).
\end{cases}
$$
You cannot use the expression of $f'_x$ to decide if $f'_x(0,0)$ exists simply because the partial derivative can exist but be discontinuous, in which case you cannot infer the value of $f'_x(0,0)$ from the expression elsewhere.
