Second order partial of $f(x,y)=\frac{xy(x^2-y^2)}{x^2+y^2}$ Consider the function $f(x,y)=\dfrac{xy(x^2-y^2)}{x^2+y^2}$ for $(x,y) \neq (0,0)$, $f=0$ otherwise.
I have to compute $\dfrac{d^2f}{dydx}(0,0)$.
I know that I have to calculate $\frac{df}{dx}$ first.
But that is ,
$\frac{df}{dx} = \frac{\partial f}{\partial y}\frac{dy}{dx}+\frac{\partial f}{\partial x}$.
When I put it in wolframalpha, it gives me this calculation, and also just $\frac{\partial f}{\partial x}$ as an alternative form.
wolframalpha calculation link
Why do I just ignore the y'? I don't know what y(x) is, as a function of x. 
y and x are independent functions, no?
 A: For $(x,y)\ne(0,0)$ one easily computes
$$f_x(x,y)={x^4y + 4x^2y^3-y^5\over (x^2+y^2)^2}\ .$$
The point $(0,0)$ is special; therefore we have to resort to the definition of $f_x(0,0)$:
$$f_x(0,0)=\lim_{x\to0}{f(x,0)-f(0,0)\over x}=0\ .$$
It follows that
$$f_x(0,y)=\cases{-y&$(y\ne0)$\cr 0&$(y=0)$\cr}\ ,$$
whence $f_x(0,y)=-y$ for all $y$. This leads to
$${\partial^2 f \over\partial y\partial x}(0,0)=-1\ .$$
A similar computation would give
$${\partial^2 f \over\partial x\partial y}(0,0)=1\ ,$$
the reason being that these mixed derivatives both fail to be continuous at $(0,0)$.
A: Given 
$$f(x,y) = \frac{{xy({x^2} - {y^2})}}{{{x^2} + {y^2}}}$$
The graph looks like:

Got it! 
For derivative with respect to x-Axis we have to consider:
$$\frac{{f(h,y) - f(0,y)}}{h} = \frac{{y\left( {{h^2} - {y^2}} \right)}}{{{h^2} + {y^2}}}$$
Then we get:
$$\mathop {\lim }\limits_{h \to 0} \frac{{f(h,y) - f(0,y)}}{h} =  - y$$
If we instead consider derivative with respect to y-Axis, we are using:
$$\frac{{f(x,h) - f(x,0)}}{h} = \frac{{x\left( {{x^2} - {h^2}} \right)}}{{{h^2} + {x^2}}}$$
An in the limit, we get:
$$\mathop {\lim }\limits_{h \to 0} \frac{{f(x,h) - f(x,0)}}{h} = x$$
Now the mixed partials must be different.
