Answer
$f_x=2 \sin(x-3y) \cos(x-3y)$
and $f_y=-6 \sin(x-3y) \cos(x-3y)$
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.
$f_x=2 \sin(x-3y) \cos(x-3y)$
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.
$f_y=2 \sin(x-3y) \cos(x-3y)\times (-3)=-6 \sin(x-3y) \cos(x-3y)$
