Are we unable to find the extrema of the function $f(x,y) = \frac{y}{x^2+y^2}$ using the second partial test? 
$$\begin{align*}
&f(x,y) = \frac{y}{x^2+y^2}\\
&f_{xx} = \frac{∂}{∂x}\left(-\frac{2xy}{(x^2+y^2)^2}\right)=-\frac{2y(y^2-3x^2)}{(x^2+y^2)^3}\\
&f_{yy} = \frac{∂}{∂x}\left(\frac{x^2-y^2}{(x^2+y^2)^2}\right)=\frac{2y(y^2-3x^2)}{(x^2+y^2)^3}
\end{align*}$$
$a$ and $b$ must be $(0,0)$ or at least one of them has to be $0$ to be critical points of the function
if the point is $(0,0)$, $f_{xx}$ or $f_{yy}$ will be undefined and finding the point $(?,0)$ for which $f_x$ or $f_y$ would $= 0$ would be mighty hard. 
Are we unable to find the extrema of this function using the second partial test?

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.
