Баxa вот. dz/dx=2xy/(x^2-y^2), dz/dy=ln(x^2-y^2)-2y^2/(x^2-y^2). 1/x*dz/dx=2y/(x^2-y^2), 1/y*dz/dy=1/y*ln(x^2-y^2)-2y/(x^2-y^2), 1/x*dz/dx+1/y*dzdy=1/y*ln(x^2-y^2)=1/y^2*yln(x^2-y^2)=z/y^2