Showing that partial derivatives of $f(x) = \frac{xy}{x^2+y^2}$ exist for $(x,y) \neq (0,0)$ I am aware there are many questions similar to this on MSE but I am having trouble following any of the solutions given. 
I have the function given by 
$f(x) = \dfrac{xy}{x^2+y^2}\;$ for $(x,y)\neq (0,0)$ and $f(0,0)=0$
I have calculated the partial derivatives and found: 
$$\frac{\partial f}{\partial x} = \frac{y(-x^2+y^2)}{(x^2+y^2)^2}, \quad \frac{\partial f}{\partial y} = \frac{x(-y^2+x^2)}{(y^2+x^2)^2}$$
Now I need to show the partial derivatives exist for $(x,y) = (0,0)$. 
I am also asked to show it is not continuous at $(x,y)=(0,0)$, based on other answers I've seen it seems like this follows from solving the first part but I fail to see how that follows through as well. 
I could just copy the solutions with my function in place as I have seen a lot of answers using the definition of the derivative, but realistically I do want to understand what the thinking behind this is. 
Any help would be appreciated. 
 A: Why we can use the definition is pretty obvious - if something does not satisfy the defining criteria of some property then that thing does not possess the property - this is the nature of definitions. Why we should use the definition, well who's to say. If consideration of a definition leads to an easy answer to a question we have, why not consider it?
Consider the quotient definition for partial differentiation: let $\mathbf{0}=(0,0)$ and $\mathbf{e}_1,\mathbf{e}_2$ be the standard basis for the plane. For any nonzero $t \in \mathbb{R}$ we have
$$Q(t) := \frac{f(\mathbf{0} + t\mathbf{e_1}) - f(\mathbf{0})}{t} = \frac{f(t,0) - f(0,0)}{t} = 0.$$
This is true because $f(t,0) = f(0,0) = 0$ for any such $t$. By definition $$\frac{ \partial f}{ \partial x}(\mathbf{0}) := \lim_{t\to 0} Q(t) = 0.$$
Similarly,
$$\frac{ \partial f}{\partial y}(\mathbf{0}) := \lim_{t\to 0} \frac{f(\mathbf{0}+ t\mathbf{e}_2) - f(\mathbf{0})}{t} = \lim_{t\to 0}\frac{f(0,t) - f(0,0)}{t} $$
$$= \lim_{t \to 0}\frac{0 - 0}{t} =\lim_{t \to 0}\frac{0}{t} = \lim_{t\to 0} 0 = 0. $$
As for continuity of $f$ at the origin, if it were true that $f$ is continuous there, then for any sequence $\mathbf{x}_n \to \mathbf{0}$ we would have $\lim_{n \to \infty} f(\mathbf{x}_n) = f(\mathbf{0}) = 0$. But the sequence $\mathbf{x}_n = \left(\frac{1}{n},\frac{1}{n}\right) \to \mathbf{0}$ is such that $f(\mathbf{x}_n) = 1/2$ for every index $n$. It certainly isn't true then that $f(\mathbf{x}_n) \to \mathbf{0}$, and we deduce that $f$ cannot be continuous at the origin.
A: The two questions about partial derivatives and continuity are independent.
For partial derivatives, you can use the definition, as in other answer, but you can also think geometrically.
The partial derivative $\frac{\partial f}{\partial x}$ at $(0,0)$ is the derivative of the curve that results from the intersection of the surface and the vertical plane $y=0$. In other words, perform the substitution $y=0$ and you obtain the 1-dimensional function: $g(x)=f(x,0)=0$. Therefore,
$$\frac{\partial f}{\partial x}(0,0)=g'(0)$$
And, yes, the function $g$ seems pretty derivable.
As for continuity, the limit does not exist, e.g. by comparing trajectories $x=0$ and $x=y$.
