Prove $gF$ is a conservative vector field if and only if $\nabla g$ is parallel to $F$ everywhere What I am given in the problem: 
$F(x,y) = (P(x, y), Q(x, y))$ is a conservative vector field defined on $\mathbb{R}^2$
$P$ and $Q $ are smooth functions, and $F \neq(0, 0)$ for all points
Problem: if $g:\mathbb{R}^2 \to \mathbb{R}$ is a smooth function, prove that $gF$ is conservative if and only if $\nabla g$ is parallel to F everywhere.
I know that since $F$ is conservative and $P$ and $Q $ are smooth functions, then $\frac{\delta Q}{\delta x}=\frac{\delta P}{\delta y} $ and that there exists an $f$ such that $\nabla f = F$. Should I assume that the given $g$ is the potential function of F ? 
Any help would be greatly appreciated!
 A: (I'm reading your $\Delta$ as $\nabla$.)
Since both ${\bf F}$ and $g$ are defined in all of ${\mathbb R}^2$ it is sufficient to look at
$${\rm curl}(g\,{\bf F})=(gQ)_x-(gP)_y=g(Q_x-P_y)+(g_xQ-g_yP)\ .$$
Here the first part on the RHS vanishes by assumption on ${\bf F}$. The second part
$$g_xQ-g_yP={\rm det}\left[\matrix{g_x&g_y\cr P&Q\cr}\right]$$
vanishes iff the vectors $\nabla g=(g_x,g_y)$ and $(P,Q)\ne{\bf 0}$ are linearly dependent, and this is the case iff $\nabla g(x,y)=\lambda\, {\bf F}(x,y)$. Since we want this to be true for all $(x,y)\in{\mathbb R}^2$ the claim follows.
A: If $gF$ is conservative, then
$\nabla \times (gF)= 0; \tag 1$
we have
$\nabla \times (gF) = \nabla \times (gP, gQ) = \dfrac{\partial (gQ)}{\partial x} - \dfrac{\partial (gP)}{\partial y} = (gQ)_x - (gP)_y, \tag 2$
where we have introduced subscript notation for partial derivatives in the right-most equality in (2).  
We further have
$(gQ)_x - (gP)_y = g_x Q + gQ_x - g_yP - gP_y; \tag 3$
then from (1) and (2),
$(gQ)_x - (gP)_y = g_x Q + gQ_x - g_yP - gP_y = 0, \tag 4$
or
$g_xQ - g_yP = gP_y - gQ_x = g(P_y - Q_x) = -g \nabla \times (P, Q) = -g \nabla \times F = 0, \tag 5$
since $F = (P, Q)$ is itself conservative, i.e.,
$\nabla \times F = 0; \tag 6$
thus
$g_xQ - g_y P = 0; \tag 7$
now if
$\nabla g = 0, \tag 8$
then (7) trivially holds; I suppose this may be considered a special case of co-linearity since then
$\nabla g = 0 (P, Q); \tag 9$
having dealt with this eventuality, we may henceforth assume that
$\nabla g \ne 0; \tag{10}$
since
$F = (P, Q) \ne 0, \tag{11}$
at least one of $P, Q \ne 0$ at each point $(x, y)$; suppose then for the moment that
$P \ne 0; \tag{12}$
then from (7),
$g_y = \dfrac{Q}{P} g_x; \tag{13}$
thus
$\nabla g = (g_x, g_y) = (g_x, \dfrac{Q}{P} g_x) = \dfrac{g_x}{P}(P, Q) = \dfrac{g_x}{P}F; \tag{14}$
in the light of (10) we see that 
$g_x \ne 0 \ne \dfrac{g_x}{P}, \tag{15}$
and thus $\nabla g$ and $F$ are co-linear in this case.  A parallel argument applies in the event that
$Q \ne 0, \tag{16}$
in which case we obtain
$\nabla g = \dfrac{g_y}{Q}F, \tag{17}$
with
$g_y \ne 0, \tag{18}$
and again, $\nabla g$ and $F$ are colinear.
