Second order partial of $f(x,y)=\frac{xy(x^2-y^2)}{x^2+y^2}$ [duplicate]

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 $(0,0)$.

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.