Answer
\begin{aligned} \frac{d y}{d x}= \frac{y^{2}-x^{2}}{2xy+2y(x^{2}+y^{2})^2}\end{aligned}
Work Step by Step
Given $$\frac{x}{(x^{2}+y^{2})}-y^2=6$$
$$ \Rightarrow \frac{x}{(x^{2}+y^{2})}-y^2-6=0$$
by letting $$F(x,y)= \frac{x}{(x^{2}+y^{2})}-y^2-6$$
So, we have
$$F_x(x,y)=\frac{\partial F(x,y)}{\partial x}= \frac{(1)(x^2+y^2)-(2x)(x)}{(x^{2}+y^{2})^2} \\
\quad \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{-x^{2}+y^{2}}{(x^{2}+y^{2})^2} $$
$$F_y(x,y)=\frac{\partial F(x,y)}{\partial y}= \frac{0-(2y)(x)}{(x^{2}+y^{2})^2}-2y \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{-2xy-2y(x^{2}+y^{2})^2}{(x^{2}+y^{2})^2}$$
Also, we get
\begin{aligned} \frac{d y}{d x}=&-\frac{F_{x}(x, y)}{F_{y}(x, y)} \\
&=-\frac{\frac{ -x^{2}+y^{2}}{ (x^{2}+y^{2})^2}}{\frac{ -2xy-2y(x^{2}+y^{2})^2}{ (x^{2}+y^{2})^2}} \\ &=-\frac{-x^{2}+y^{2}}{-2xy-2y(x^{2}+y^{2})^2}\\
&=\frac{y^{2}-x^{2}}{2xy+2y(x^{2}+y^{2})^2}\end{aligned}
