Find $f_x,f_y,f_x(-4,1)$, and $f_y(-1,-7)$ for the following equation.
$$f(x,y)=\sqrt{x^2+y^2}$$

Step 1 ：Given the function $f(x,y)=\sqrt{x^2+y^2}$, we are asked to find the partial derivatives of the function with respect to $x$ and $y$, and then to evaluate these 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 will differentiate $f(x,y)$ with respect to $x$, treating $y$ as a constant. Similarly, to find $f_y$, we will differentiate $f(x,y)$ with respect to $y$, treating $x$ as a constant.

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

Step 5 ：Evaluating these at the given points, we find $f_x(-4,1)=\frac{-4}{\sqrt{17}}$ and $f_y(-1,-7)=\frac{-7}{\sqrt{50}}$.

Step 6 ：So, the final answers are \(\boxed{f_{x} = \frac{x}{\sqrt{x^{2}+y^{2}}}\), \(\boxed{f_{y} = \frac{y}{\sqrt{x^{2}+y^{2}}}\), $\boxed{{\displaystyle f_x(-4,1)=\frac{-4}{\sqrt{17}}}}$, and $\boxed{{\displaystyle f_y(-1,-7)=\frac{-7}{\sqrt{50}}}}$.