Answer
$f_x=6(2x-3y)^2$ and $f_y=-9(2x-3y)^2$
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, and vice versa:
$f_x=3(2x-3y)^2\times 2 \\f_y=3(2x-3y)^2\times -3$
Our results are: $f_x=6(2x-3y)^2$ and $f_y=-9(2x-3y)^2$
