Answer
$w_x=y\sec^2{(x+2z)}$, $w_y=\tan{(x+2z)}$, $w_z=2y\sec^2{(x+2z)}$.
Work Step by Step
$w=y\tan{(x+2z)}$
In order to find $w_x$ we treat $y$ and $z$ as constants and differentiate with respect to $x$.
$w_x=y\sec^2{(x+2z)}$
Analogously
$w_y=\tan{(x+2z)}$
$w_z=2y\sec^2{(x+2z)}$