I have a Field $F(x,y):= \left( \begin{array}{c} ay\\ 0\\ \end{array} \right)$ and I have to find out whether this field is potential. Well despite the terminology( wich is not completely clear to me) i thought that the answer is affermative iff there exist a $U$ such that $F=-\nabla U$. An easy way would be to check if
$\dfrac{\partial F_x}{\partial y} = \dfrac{\partial F_y}{\partial x}$ holds. ( as suggested on https://physics.stackexchange.com/)
But here is my approach ( and here the question is purely mathematical) :
(1) $ay=\dfrac{\partial U}{\partial x} \Leftrightarrow \int \partial U= \int ay \partial x \Leftrightarrow U=axy + C(y)$ , where $C$ is a function of $y$ only.
(2) $0=\dfrac{\partial U}{\partial y}=ax+\dfrac{\partial C(y)}{\partial y}$ (subst. (1) in (2)) $\Leftrightarrow \partial C(y)=-ax \partial y$ but here we can clearly see that $C(y)$ depends on $x$ , wich is not possible , so there can be no potential $U$.
Is this approach right? Is the mathematics behind it right? (that's one of the first such problems that i'm solving please be patient)
Let me try to understand you first as the question is quite confusing; given a field:
$$\mathbf F(y):= \left(\begin{array}{c}ay\\0\\\end{array}\right)$$
first we need to confirm a conservative vector field:
$$\dfrac{\partial F_1}{\partial y} - \dfrac{\partial F_2}{\partial x}\overset{?}{=}0$$
with
$$\dfrac{\partial F_1}{\partial y}=a$$ $$\dfrac{\partial F_2}{\partial y}=0$$
Hence not conservative. This is precondition to any further calculation that would make sense and results in potential not existing.
Now you take the tour and claim, if there would be a potential:
$$\dfrac{\partial U}{\partial x} =ay \rightarrow U\sim \int dU= \int a\,y \,dx=a\,y\,x+C_1(y)$$
similarly:
$$\dfrac{\partial U}{\partial y} =0 \rightarrow U \sim \int dU= C_2$$
And you have a contradiction because the precondition is not fulfilled.
Resume: Not every conservative field that fulfills the abovementioned precondition has a potential but every who has a potential will be corresponding to a conservative field.
To your questions: yes your approach and calculation is absolutely correct.
To your comment, I see it like this: If $\mathbf F$ is a conservative field, then there exists some potential function $U$ so that $\nabla U= \mathbf F$.