If $F(x,y)=f(x) g(y)$ and the required derivatives exist, just compute the required partials.
Note these are partial derivatives. When finding $F_x(x,y)$, for example, you think of the variable $y$ as fixed and differentiate with respect to $x$: F_x(x,y) ={\textstyle{\partial \over\partial x}}\bigl[\, f(x)g(y)\,\bigr] = g(y) {\textstyle{\partial \over\partial x}} f(x) =g(y){\textstyle{d\over dx} }f(x)= g(y)f'(x).
You have:
\ \ \ \ F_x(x,y)=f'(x) g(y),
\ \ \ \ F_y(x,y)=f(x)g'(y),
and
\ \ \ \ F_{xy}(x,y)={\partial\over\partial y}F_x(x,y) ={\partial\over\partial y} \bigl[\,f'(x)g(y)\,\bigr] =f'(x)g'(y).
Now substitute into your equation and show that it's true.
Remark: Note you reversed the order of differentiation in your expression for $F_{xy}=(F_x)_y$; you find $F_x$ first. (It is not always the case that $F_{xy}=F_{yx}$.)