Answer
$f_x=2sin(x-3y)cos(x-3y)$
$f_y=-6sin(x-3y)cos(x-3y)$
Work Step by Step
Take the first partial derivatives of the given function. When taking a partial derivative with respect to x, treat y as a constant, and vice versa:
$f_x=2sin(x-3y)cos(x-3y)$
$f_y=2sin(x-3y)cos(x-3y)\times-3=-6sin(x-3y)cos(x-3y)$
