Find $f_{x}, f_{y}, f_{x}(-6,-5)$, and $f_{y}(1,-2)$ for the following equation.
\[
f(x, y)=\sqrt{x^{2}+y^{2}}
\]
\[
\mathrm{f}_{\mathrm{x}}=\square
\]
(Type an exact answer, using radicals as needed.)

Step 1 ：Given the function \(f(x, y)=\sqrt{x^{2}+y^{2}}\), we need to find the partial derivatives \(f_{x}\) and \(f_{y}\), and evaluate them at specific points.

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 \(f_{x}\), we differentiate \(f(x, y)\) with respect to \(x\), treating \(y\) as a constant. Similarly, to find \(f_{y}\), we differentiate \(f(x, y)\) with respect to \(y\), treating \(x\) as a constant.

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

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

Step 6 ：To find \(f_{x}(-6,-5)\), we substitute the point (-6,-5) into \(f_{x}\) to get \(f_{x}(-6,-5) = \frac{-6}{\sqrt{61}}\).

Step 7 ：To find \(f_{y}(1,-2)\), we substitute the point (1,-2) into \(f_{y}\) to get \(f_{y}(1,-2) = \frac{-2}{\sqrt{5}}\).

Step 8 ：Final Answer: \(f_{x} = \boxed{\frac{x}{\sqrt{x^2 + y^2}}}\), \(f_{y} = \boxed{\frac{y}{\sqrt{x^2 + y^2}}}\), \(f_{x}(-6,-5) = \boxed{\frac{-6}{\sqrt{61}}}\), \(f_{y}(1,-2) = \boxed{\frac{-2}{\sqrt{5}}}\).