= Solution
For an <exact differential form>, equality of the mixed <partial derivatives> requires $P_y=Q_x$. Here
$$
P_y=\frac{y}{a\sqrt x\sqrt{1+y^2}},\qquad Q_x=\frac{y}{2\sqrt x\sqrt{1+y^2}}.
$$
Equality for all $x>0$ and real $y$ forces $1/a=1/2$, so $\boxed{a=2}$. To construct the potential, integrate $P$ with respect to $x$:
$$
v(x,y)=\sqrt x\left(1+\sqrt{1+y^2}\right)+C(y).
$$
Differentiation with respect to $y$ gives $v_y=Q+C'(y)$, so $C'(y)=0$. Thus the most general potential on the connected domain $x>0$ is
$$
\boxed{v(x,y)=\sqrt x\left(1+\sqrt{1+y^2}\right)+C,\qquad C\in\mathbb R.}
$$
Its explicit <partial derivatives> verify sufficiency as well as necessity.
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