$u_x\+u_y$

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

$\=\left(f_r \left(1\right)\+f_s \left(-1\right)\right)\+\left(f_r \left(-1\right)\+f_s \left(1\right)\right)$

$\=\left(1-1\right) f_r\+\left(-1\+1\right) f_s$

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

$\=0\+0$

$\=0$

Maple's statement of the general chain rule for this case

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

${\mathrm{D}}_1\left(f\right)\left(r\left(x\,y\right)\,s\left(x\,y\right)\right)\left(\frac{\partial }{\partial x}r\left(x\,y\right)\right)\+{\mathrm{D}}_2\left(f\right)\left(r\left(x\,y\right)\,s\left(x\,y\right)\right)\left(\frac{\partial }{\partial x}s\left(x\,y\right)\right)\+{\mathrm{D}}_1\left(f\right)\left(r\left(x\,y\right)\,s\left(r\,y\right)\right)\left(\frac{\partial }{\partial y}r\left(x\,y\right)\right)\+{\mathrm{D}}_2\left(f\right)\left(r\left(x\,y\right)\,s\left(r\,y\right)\right)\left(\frac{\partial }{\partial y}s\left(r\,y\right)\right)$

Specify the arguments of $f$

$\frac{\partial }{\partial x} f\left(x-y\,y-x\right)\+\frac{\partial }{\partial y} f\left(x-y\,y-x\right)$ = $0$

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

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

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

Apply the chain rule to obtain $u_x\+u_y$ 

Calculus palette: Partial-differential operator

Context Panel: Evaluate and Display Inline

$\frac{\partial }{\partial x} f\left(r\,s\right)\+\frac{\partial }{\partial y} f\left(r\,s\right)$ = $0$

Implement the chain rule

$\mathrm{diff}\left(f\left(r\,s\right)\,x\right) \+\mathrm{diff}\left(f\left(r\,s\right)\,y\right)$ = $0$