Finding unknown function in conservative field 
So here $M(x,y)=8x+3y$
and
$N(x,y)=g(x,y)+18y^2$
$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}=3$
Integrating $\frac{\partial N}{\partial x}$ with respect to $x$ and holding $y$ fixed
$N(x,y)=\int\frac{\partial N}{\partial x}=3x+h(y)$
differentiating $N(x,y)$ with respect to $y$
$\frac{\partial N}{\partial y}=\frac{\partial g}{\partial y}+36y$
I've noticed that however much i try integrating or differentiating either $M(x,y)$ or $N(x,y)$ and their partial derivatives, i always end up with two unknowns. Can someone help?
 A: As all integrals of $\mathbf{F}$ around loops are $0$ on (the open, simply-connected) plane, you are right that $\mathbf{F}$ is conservative.  (I'm assuming more than mere continuity here.)  You are right in asserting $M_y = N_x = 3$ then follows.  Since $\frac{\partial}{\partial x} 18y^2=0$ then $g(x,y)$ may take the form $g(x,y) = 3x + h(y)$ for any function $h$ of the single variable $y$.  
Did anyone assert that the solution for the function $g$ had to be unique?
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I think it's actually okay that there are some unknowns.
The reason is there are many possible $g(x,y)$ functions that will solve this problem, i.e. that will satisfy the condition that the resulting vector field is conservative.
You correctly stated that
$$ \frac{\der M}{\der y} = \frac{\der N}{\der x} $$
which simplifies to
$$ 3 = g_x $$
Integrating both sides with respect to $x$ yields
$$ g(x,y) = 3x + h(y) $$
where $h(y)$ is an arbitrary function.
At this point you might be tempted to try to somehow solve for $h(y)$ (I was, anyway), but it turns out this $h(y)$ is not something to solve for, it rather describes the form of all the possible $g(x,y)$ solutions: i.e. that as long as $g(x,y)$ has the form
$$ g(x,y) = 3x + h(y) $$
for some function $h(y)$, then $g$ is a valid solution to the problem. If you don't believe me, try plugging in the above expression for $g(x,y)$ and solving for the potential function $f$ of the vector field $\vec{\mathbf{F}}$ in terms of $h(y)$. You'll see that it doesn't matter what $h(y)$ is, we can always find $f$.
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