Answer
$(\sqrt 3,\frac{7\pi}{4},0.955)$.
Work Step by Step
Since $ x=1, \, y=-1, \, z=1$ and
$$\rho=\sqrt{x^2+y^2+z^2},\quad \theta=\tan^{-1}( y/x), \quad \phi=\cos^{-1}(z/\rho),$$
then we have
$$\rho=\sqrt{1+1+1 }= \sqrt 3, $$
$$\theta=\tan^{-1}(-1/1)=\frac{7\pi}{4}, $$
$$\phi=\cos^{-1}(1/\sqrt 3)=0.955.$$
Hence, the spherical coordinates are $(\sqrt 3,\frac{7\pi}{4},0.955)$.
