Answer
$f_x=1$
$f_y=-\dfrac{y}{\sqrt {y^2+z^2}}$
$f_z=-\dfrac{z}{\sqrt {y^2+z^2}}$
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 $y$ and $z$ as a constant, and vice versa:
$f_x=1$
$f_y=-\dfrac{1}{2} (y^2+z^2)^{-1/2}\times 2y=-\dfrac{y}{\sqrt {y^2+z^2}}$
$f_z=-\dfrac{1}{2}(y^2+z^2)^{-1/2}\times 2z=-\dfrac{z}{\sqrt {y^2+z^2}}$
