Given a vector field ${\bf F}(x, y)$, find a function $h(x)$ such that $h {\bf F}$ is conservative 
Let $${\bf F}(x,y)=y{\bf i}-6x{\bf j}.$$ Find a nonzero function $h(x)$, where $h(1)=5$, such that $h(x) {\bf F}(x,y)$ is a conservative vector field.

I know that this is essentially asking for a function $h(x)$ that when multiplied by ${\bf F}(x, y)$ is equal to the gradient of some other potential function. I'm having trouble, however, coming up with a function $h(x)$ for which $h(1)=5$. Any help would be great, thanks in advance!
 A: $hF$ is conservative if and only if
$\nabla \times (hF) = 0; \tag 1$
that is,
$\nabla h \times F + h\nabla \times F = 0, \tag 2$
acording to a well-known vector identity; we have
$F = y \mathbf i - 6x \mathbf j, \tag 3$
$\nabla \times F = -7\mathbf k, \tag 4$
$\nabla h = h_x \mathbf i, \tag 5$
$\nabla h \times F = -6x h_x \mathbf k; \tag 6$
from (2), (4) and (6),
$-6xh_x = 7h; \tag 7$
$(\ln h)_x = -\dfrac{7}{6x}; \tag 8$
$\ln \left (\dfrac{h(x)}{h(1)} \right ) = \ln h(x) - \ln h(1) = \displaystyle \int_1^x (\ln h)_s \; ds = -\int_1^x \dfrac{7}{6s}  \; ds =-\dfrac{7}{6} \ln x; \tag 9$
$\dfrac{h(x)}{h(1)} = x^{-7/6}: \tag{10}$
$h(x) = h(1) x^{-7/6} = 5x^{-7/6}.  \tag{11}$
Check: from (11),
$h_x(x) = -\dfrac{35}{6} x^{-13/7}; \tag{12}$
$-6xh_x = 35 x^{-6/7} = 7 (5x^{-6/7}) = 7h, \tag{13}$
validating (7); substituting this into (6) we obtain 
$\nabla h \times F = 7h \mathbf k; \tag{14}$
from (4), 
$h\nabla \times F = -7h\mathbf k; \tag{15}$
scrutinizing (14) and (15), we see that (2) binds, hence also (1).
A: This method is morally equivalent to Robert's very good answer, but the presentation is deliberately different.
If $f(x, y)$ is a potential function for $h(x) {\bf F}(x, y)$ then (suppressing arguments and) comparing the components of $$\nabla f = h {\bf F} = h (y {\bf i} - 6 x {\bf j})$$ gives the system
\begin{align*}
f_x &= y h \\
f_y &= -6 x h . \\
\end{align*}
Differentiating and applying Clairaut's Theorem (which in particular means that we assume $f$ is $C^2$) gives the differential equation
$$h = f_{xy} = f_{yx} = -6 (h + x h') .$$
Rearranging gives a separable o.d.e.,
$$\frac{7}{x} = -\frac{6 h'}{h} .$$

 Integrating both sides and solving for $h$ gives $h(x) = C |x|^{-7 / 6}$, and it remains to solve for $C$ using the initial condition $h(1) = 5$. Note, however, that $h$ is not even defined for all $x$, and so it only defines a conservative vector field $h {\bf F}$ on, e.g., the right half-plane, $\{x > 0\}$, and not on all of the presumptive domain $\Bbb R^2$ of $\bf F$.

