$y F_x\+x F_y$

$\=y \left(f_r r_x\+f_s s_x\right)\+x \left(f_r r_y\+f_s s_y\right)$

$\=y \left(f_r \left(2 x\right)\+f_s \left(-2 x\right)\right)\+x \left(f_r \left(-2 y\right)\+f_s \left(2 y\right)\right)$

$\=\left(f_r \left(2 x y\right)-f_s \left(2 x y\right)\right)\+\left(f_r \left(-2 x y\right)\+f_s \left(2 x y\right)\right)$

$\=\left(2 x y-2 x y\right) f_r\+\left(-2 x y\+2 x y\right) f_s$

$\=0\cdot f_r\+0\cdot f_s$

$\=0\+0$

$\=0$

Define $r\left(x\,y\right)$ and $s\left(x\,y\right)$ 

$r\=x^2-y^2$$\stackrel{\text{assign}}{\to }$

$s\=y^2-x^2$$\stackrel{\text{assign}}{\to }$

Apply the chain rule to obtain $x{F}_x\+y{F}_y$ 

$y \frac{\partial }{\partial x} f\left(r\,s\right)\+x \frac{\partial }{\partial y} f\left(r\,s\right)$

$y\left(2{\mathrm{D}}_1\left(f\right)\left(x^2-y^2\,-x^2\+y^2\right)x-2{\mathrm{D}}_2\left(f\right)\left(x^2-y^2\,-x^2\+y^2\right)x\right)\+x\left(-2{\mathrm{D}}_1\left(f\right)\left(x^2-y^2\,-x^2\+y^2\right)y\+2{\mathrm{D}}_2\left(f\right)\left(x^2-y^2\,-x^2\+y^2\right)y\right)$

$\stackrel{\text{simplify}}{\=}$

$0$

Implement the chain rule

$y \mathrm{diff}\left(f\left(r\,s\right)\,x\right) \+x \mathrm{diff}\left(f\left(r\,s\right)\,y\right)$

$y\left(2{\mathrm{D}}_1\left(f\right)\left(x^2-y^2\,-x^2+y^2\right)x-2{\mathrm{D}}_2\left(f\right)\left(x^2-y^2\,-x^2+y^2\right)x\right)+x\left(-2{\mathrm{D}}_1\left(f\right)\left(x^2-y^2\,-x^2+y^2\right)y+2{\mathrm{D}}_2\left(f\right)\left(x^2-y^2\,-x^2+y^2\right)y\right)$

$\mathrm{simplify}\left(y \mathrm{diff}\left(f\left(r\,s\right)\,x\right) \+x \mathrm{diff}\left(f\left(r\,s\right)\,y\right)\right)$

$0$