First, you’ve some errors in your third line of calculations. The correct partials are:
$\begin{align*} &f_x(x,y)=-\frac{2xy}{(x^2+y^2)^2}\\ &f_y(x,y)=\frac{x^2-y^2}{(x^2+y^2)^2}\\ &f_{xx}(x,y)=\frac{2y(3x^2-y^2)}{(x^2+y^2)^3}\\ &f_{xy}(x,y)=\frac{2x(3y^2-x^2)}{(x^2+y^2)^3}\\ &f_{yy}(x,y)=\frac{3y(y^2-2x^2)}{(x^2+y^2)^3} \end{align*}$
Clearly $f_x(x,y)=0$ only if $xy=0$, i.e., at least one of $x$ and $y$ is $0$. Similarly, $f_y(x,y)=0$ only if $0=x^2-y^2=(x-y)(x+y)$, i.e., only if $y=x$ or $y=-x$. The only way to satisfy both of these conditions is to have $x=y=0$, and the function and its partial derivatives aren’t defined at $(0,0)$. Thus, you’re quite right: there is no point at which the second derivative test applies.
Rewriting the function in polar coordinates as
$f(r,\theta)=\frac{r\sin\theta}{r^2}=\frac{\sin\theta}r$
may help to explain what’s going on. As we travel around the circle $C_r$ of radius $r$ centred at the origin, the function value is $0$ where $C_r$ crosses the $x$-axis, reaches a maximum of $\frac1r$ where $C_r$ crosses the positive $y$-axis, and reaches a minimum of $-\frac1r$ where $C_r$ crosses the negative $x$-axis. But that high point with value $\frac1r$ on $C_r$ can’t be a local maximum of the function, because the value of the function gets larger as you move down the $y$-axis towards the origin: $\frac1r$ increases as $r$ decreases. Similarly, the low point on $C_r$ can’t be a local minimum of $f$, because $-\frac1r$ gets smaller (more negative) as you move up the negative $y$-axis and $r$ decreases.