Answer
$f_x=\dfrac{1}{x+2y+3z} ; \\
f_y= \dfrac{2}{x+2y+3z} ; \\
f_z=\dfrac{3}{x+2y+3z}$
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:
$f_x=\dfrac{1}{x+2y+3z} ; \\
f_y=(2) \times \dfrac{1}{x+2y+3z}=\dfrac{2}{x+2y+3z} ; \\
f_z=(3) \times \dfrac{1}{x+2y+3z}=\dfrac{3}{x+2y+3z}$