Proving Schwarz's theorem for reversal of order of derivative 
Show that the function $f(x,y)=xy(x^2-y^2)/(x^2+y^2)$ at $(x,y) \neq (0,0)$ and equal to $0$ when $(x,y)= (0,0)$ does not satisfy conditions of Schwarz's theorem.

Here I am able to find out $f_x(x,y)$ but how to find $f_{yx}(x,y)$ 
 A: Let's take the partial derivative with respect to $x$ at $(x, y) = (0, 0)$ first. We calculate:
\begin{align*}
f_x(0, 0) &= \lim_{h \rightarrow 0} \frac{f(0 + h, 0) - f(0, 0)}{h} \\
&= \lim_{h \rightarrow 0} \frac{f(h, 0)}{h} \\
&= \lim_{h \rightarrow 0} \frac{(h)(0)(h^2 - 0^2)}{h} = 0.
\end{align*}
Similarly for $f_y(0, 0) = 0$. At other points, we have,
\begin{align*}f_x(x,y) &= \frac{\partial}{\partial x} \frac{xy(x^2 - y^2)}{x^2 + y^2} \\
&= \frac{(x^2 + y^2)\frac{\partial}{\partial x}[xy(x^2 - y^2)] - xy(x^2 - y^2)\frac{\partial}{\partial x}[x^2 + y^2]}{(x^2 + y^2)^2} \\
&= \frac{(3yx^2 - y^3)(x^2 + y^2) - xy(x^2 - y^2)(2x)}{(x^2 + y^2)^2} \\
&= y\frac{(3x^2 - y^2)(x^2 + y^2) - 2x^2(x^2 - y^2)}{(x^2 + y^2)^2} \\
&= y\frac{x^4 + 4x^2 y^2 - y^4}{(x^2 + y^2)^2}.
\end{align*}
Therefore,
\begin{align*}
f_{xy}(0, 0) &= \lim_{h \rightarrow 0} \frac{f_x(0, 0 + h) - f_x(0, 0)}{h} \\
&= \lim_{h \rightarrow 0} \frac{f_x(0, h)}{h} \\
&= \lim_{h \rightarrow 0} \frac{1}{h} \cdot h \cdot \frac{0^4 + 4 \cdot 0^2 h^2 - h^4}{(0^2 + h^2)^2} \\
&= \lim_{h \rightarrow 0} \frac{-h^4}{(h^2)^2} = -1.
\end{align*}
I'm guessing that $f_{yx}(0, 0) = 1$, but you'll have to check for yourself.
