Answer
The solution is
$$f_y(1,2,2)=\frac{1}{6}.$$
Work Step by Step
First, find the derivative:
$$f_y=\left(\ln\frac{1-\sqrt{x^2+y^2+z^2}}{1+\sqrt{x^2+y^2+z^2}}\right)'_y=\frac{1}{\frac{1-\sqrt{x^2+y^2+z^2}}{1+\sqrt{x^2+y^2+z^2}}}=\left(\frac{1-\sqrt{x^2+y^2+z^2}}{1+\sqrt{x^2+y^2+z^2}}\right)'_y=\frac{1+\sqrt{x^2+y^2+z^2}}{1-\sqrt{x^2+y^2+z^2}}\times\\
\times\frac{(1-\sqrt{x^2+y^2+z^2})'_y(1+\sqrt{x^2+y^2+z^2})-(1-\sqrt{x^2+y^2+z^2})(1+\sqrt{x^2+y^2+z^2})'_y}{(1+\sqrt{x^2+y^2+z^2})^2}=\\
\frac{1}{1-\sqrt{x^2+y^2+z^2}}\times\\
\times\frac{(1-\sqrt{x^2+y^2+z^2})'_y(1+\sqrt{x^2+y^2+z^2})-(1-\sqrt{x^2+y^2+z^2})(1+\sqrt{x^2+y^2+z^2})'_y}{(1+\sqrt{x^2+y^2+z^2})}.$$
Now we will separately calculate
$(\sqrt{x^2+y^2+z^2})'_y=\frac{1}{2\sqrt{x^2+y^2+z^2}}(x^2+y^2+z^2)'_y=\frac{2y}{2\sqrt{x^2+y^2+z^2}}=\frac{y}{\sqrt{x^2+y^2+z^2}};$
then also:
$(1-\sqrt{x^2+y^2+z^2})'_y=-\frac{y}{\sqrt{x^2+y^2+z^2}},$
and
$(1+\sqrt{x^2+y^2+z^2})'_y=\frac{y}{\sqrt{x^2+y^2+z^2}}.$
This further gives:
$(1-\sqrt{x^2+y^2+z^2})'_y(1+\sqrt{x^2+y^2+z^2})=-\frac{y}{\sqrt{x^2+y^2+z^2}}-y$
and
$(1+\sqrt{x^2+y^2+z^2})'_y(1-\sqrt{x^2+y^2+z^2})=\frac{y}{\sqrt{x^2+y^2+z^2}}-y$.
Putting this together:
$$(1-\sqrt{x^2+y^2+z^2})'_y(1+\sqrt{x^2+y^2+z^2})-(1+\sqrt{x^2+y^2+z^2})'_y(1-\sqrt{x^2+y^2+z^2})=\\
-\frac{y}{\sqrt{x^2+y^2+z^2}}-y-\left(\frac{y}{\sqrt{x^2+y^2+z^2}}-y\right)=\\
-\frac{2y}{\sqrt{x^2+y^2+z^2}}.$$
Putting all of this into the derivatve we get:
$$f_y=\frac{1}{1-\sqrt{x^2+y^2+z^2}}\cdot\frac{\frac{-2y}{\sqrt{x^2+y^2+z^2}}}{1+\sqrt{x^2+y^2+z^2}}=\frac{2y}{(x^2+y^2+z^2-1)\sqrt{x^2+y^2+z^2}}.$$
Finally, lets calculate by direct substitution $x=1$, $y=2$, $z=2$:
$$f_y(1,2,2)=\frac{2\cdot2}{(1^2+2^2+2^2-1)\sqrt{1^2+2^2+2^2}}=\frac{4}{8\cdot3}=\frac{1}{6}.$$
