Answer
$f_x= -2 x \space e^{-(x^2+y^2+z^2)} \\f_y= -2 y \space e^{-(x^2+y^2+z^2)} \\f_z= -2 z \space e^{-(x^2+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=e^{-(x^2+y^2+z^2)}\times-2x = -2 x \space e^{-(x^2+y^2+z^2)} $
$f_y=e^{-(x^2+y^2+z^2)}\times-2y = -2 y \space e^{-(x^2+y^2+z^2)} $
$f_z=e^{-(x^2+y^2+z^2)}\times-2z = -2 z \space e^{-(x^2+y^2+z^2)} $
