Find $f_{x}, f_{y}, f_{x}(-5,-1)$, and $f_{y}(2,-2)$ for the following equation.
\[
f(x, y)=\sqrt{x^{2}+y^{2}}
\]
\[
\mathrm{f}_{\mathrm{x}}=
\]
(Type an exact answer, using radicals as needed.)
\[
f_{y}=
\]
(Type an exact answer, using radicals as needed.)
\[
f_{x}(-5,-1)=
\]
(Type an exact answer, using radicals as needed.)
\[
\mathrm{f}_{\mathrm{y}}(2,-2)=
\]
(Type an exact answer, using radicals as needed.)

Step 1 ：The function given is \(f(x, y)=\sqrt{x^{2}+y^{2}}\).

Step 2 ：The partial derivative of a function with respect to a variable is the derivative of the function with respect to that variable, treating all other variables as constants.

Step 3 ：To find \(f_x\) and \(f_y\), we need to take the derivative of \(f(x, y)\) with respect to \(x\) and \(y\) respectively.

Step 4 ：The partial derivative of \(f(x, y)\) with respect to \(x\) is \(f_{x}=\frac{x}{\sqrt{x^{2}+y^{2}}}\).

Step 5 ：The partial derivative of \(f(x, y)\) with respect to \(y\) is \(f_{y}=\frac{y}{\sqrt{x^{2}+y^{2}}}\).

Step 6 ：To find \(f_x(-5,-1)\) and \(f_y(2,-2)\), we substitute these values into the partial derivatives we found.

Step 7 ：The value of \(f_{x}\) at the point \((-5,-1)\) is \(f_{x}(-5,-1)=\frac{-5}{\sqrt{26}}\).

Step 8 ：The value of \(f_{y}\) at the point \((2,-2)\) is \(f_{y}(2,-2)=\frac{-2}{\sqrt{8}} = -\frac{\sqrt{2}}{2}\).

Step 9 ：\(\boxed{f_{x}=\frac{x}{\sqrt{x^{2}+y^{2}}}}\)

Step 10 ：\(\boxed{f_{y}=\frac{y}{\sqrt{x^{2}+y^{2}}}}\)

Step 11 ：\(\boxed{f_{x}(-5,-1)=\frac{-5}{\sqrt{26}}}\)

Step 12 ：\(\boxed{f_{y}(2,-2)=-\frac{\sqrt{2}}{2}}\)