Answer
$\frac{1}{6}$
Work Step by Step
We can change the question into a simpler form by the property of logarithm
$f(x, y, z)=\ln\biggl(\frac{1-\sqrt{x^2+y^2+z^2}}{1+\sqrt{x^2+y^2+z^2}}\biggr)=\ln(1-\sqrt{x^2+y^2+z^2})-\ln (1+\sqrt{x^2+y^2+z^2})$
$f_y=\frac{\frac{-y}{\sqrt{x^2+y^2+z^2}}}{1-\sqrt{x^2+y^2+z^2}}-\frac{\frac{y}{\sqrt{x^2+y^2+z^2}}}{1+\sqrt{x^2+y^2+z^2}}$
Then
$f_y(1, 2, 2)=\frac{-\frac{2}{3}}{1-3}-\frac{\frac{2}{3}}{1+3}=\frac{1}{6}$
