Answer
$f_x=\dfrac{x}{\sqrt {x^2+y^2}} ; \\
f_y=\dfrac{y}{\sqrt {x^2+y^2}} $
Work Step by Step
Our aim is to take the first partial derivatives of the given function $f(x,y)$ with respect to $x$, by keeping $y$ as a constant, and vice versa:
$f_x=\dfrac{1}{2} \times (x^2+y^2)^{-1/2}\times 2x ; \\
f_y=\dfrac{1}{2} \times (x^2+y^2)^{-1/2}\times 2y$
Our results are: $f_x=\dfrac{x}{\sqrt {x^2+y^2}} ; \\
f_y=\dfrac{y}{\sqrt {x^2+y^2}} $
