Answer
$-6 \sin (3x-y^2) \cos (3x-y^2) $ and $4y \sin (3x-y^2) \cos (3x-y^2) $
Work Step by Step
We need to take the first partial derivatives of the given function.
In order to find the partial derivative, we will differentiate with respect to $x$, by keeping $y$ as a constant, and vice versa:
$f_x=2 \cos (3x-y^2) [-\sin (3x-y^2) \times \dfrac{\partial (3x-y^2)}{\partial x}=-6 \sin (3x-y^2) \cos (3x-y^2) $
$f_y=2 \cos (3x-y^2) [-\sin (3x-y^2) \times \dfrac{\partial (3x-y^2)}{\partial y}=4y \sin (3x-y^2) \cos (3x-y^2) $
