Potentially good?

First of all, how to check whether a field is conservative(potential field)?

$\displaystyle\oint_C \vec{F} \mathrm{d}\vec{r} = 0$

Stoke's theorem states that

$\displaystyle\oint_C \vec{F} \mathrm{d}\vec{r} = \int_S (\nabla \times \vec{F})\: \mathrm{d}\vec{a}$

So if $(\nabla \times \vec{F}) = 0$ , we can say a force to be conservative.

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$\displaystyle\nabla= \frac{\partial}{\partial x} \hat{\imath} + \frac{\partial}{\partial y} \hat{\jmath} + \frac{\partial}{\partial z} \hat{k}$

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Let's take $\displaystyle 1^{\mathrm{st}}$ Force,

$\displaystyle\vec{F_1} = ay\hat{\imath}$

$\begin{aligned}\displaystyle\nabla \times \vec{F_1} &= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ ay & 0&0\end{vmatrix} \\&= \displaystyle\frac{\partial\: ay}{\partial z}\hat{\jmath} - \frac{\partial\:ay}{\partial y} \hat{k} \\&=\displaystyle -a\hat{k} \\&\neq 0\end{aligned}$

Hence $\;\vec{F_1}\;$ is not a potential field or a conservative force.

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Let's take $\displaystyle 2^{\mathrm{nd}}$ Force,

$\displaystyle \vec{F_2} = ax\hat{\imath} + by\hat{\jmath}$

$\begin{aligned} \nabla \times \vec{F_2} & =\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ ax & by&0\end{vmatrix} \\ &=\displaystyle -\frac{\partial\: by}{\partial z} \hat{\imath} + \frac{\partial \: ax}{\partial z} \hat{\jmath} +\left(\frac{\partial\: by}{\partial x} - \frac{\partial \: ax}{\partial y}\right)\hat{k} \\&= 0\end{aligned}$

$∴\; \boxed{\vec{F_2} = ax\hat{\imath} + by\hat{\jmath}}$ is a potential field or a conservative force.