Let
$$ f(x,y) = \begin{cases} xy \frac{x^2 - y^2}{x^2 + y^2}, & (x,y) \ne (0,0) \\ 0, & (x,y) = (0,0) \end{cases} $$
Compute $f_x (0,0)$, $f_y (0,0)$, $f_{xx} (0,0)$, $f_{xy} (0,0)$, and determine where it is continuous.
Here's what I did
First, I simplified the equation:
$$ f(x,y) = xy \frac{x^2 - y^2}{x^2 + y^2} = \frac{x^3y - xy^3}{x^2 + y^2}$$
Recallinig the definition of a partial derivative:
$$f_x = lim_{h \to 0} \frac{f(x+h,y) - f(x,y)}{h}$$
I find that
$$f_x(0,0) = lim_{h \to 0} \frac{\frac{(x+h)^3y - (x+h)y^3}{(x+h)^2 + y^2} - 0}{h} = 0$$
Because
$$ lim_{h \to 0} \frac{x^3y + h(...)y - xy^3 - h(y^3)}{(x^2 + h(...) + y^2)h}$$ becomes zero when $(x,y) = (0,0)$, right?
But now my concern is repeating this for $f_{xx}$. Do I repeat the operation with $f_x$ being the new $f(x,y$? Using the definition as a method? At the end I don't know if it will become 0 again. I tried it before but it becomes such a long and tedious expansion of terms.
I can't see the end in sight.
Why can't you convert it in to polar co-ordinate
Put $x=rcos(\theta)$ and $ y=rsin(\theta)$ then on simplification you will get $f(r,\theta) = \frac {r^2}{4}sin(4\theta)$ now it will be very easy to differentiate now