I don't have idea for do this. I need a hint or something, someone could you help me?
If the $\mathbf{k}$ component, $ \frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}$ of the curl of $\mathbf{\bar{F}}$ is positive everywhere, use Green’s Theorem to show that there is a closed curve $\mathit{C}$ such that $\oint_C \mathbf{\bar{F}}\cdot d \mathbf{\bar{r}}\neq \mathbf{\bar{0}}$
If the $\mathbf{j}$ component, $ \frac{\partial F_{1}}{\partial z} - \frac{\partial F_{3}}{\partial x}$ of the curl of $\mathbf{\bar{F}}$ is positive everywhere, show that there is a closed curve $\mathit{C}$ such that $\oint_C \mathbf{\bar{F}}\cdot d \mathbf{\bar{r}}\neq \mathbf{\bar{0}}$
Let $C$ be a closed curve bounding some region $D$ in the $xy$-plane of $\mathbb{R}^3$. We have $$ \oint_C \mathbf{F} \cdot d\mathbf{r} = \oint_C F_1\ dx + F_2\ dy + F_3\ dz = \underbrace{\oint_C F_1\ dx + F_2\ dy}_{a} + \underbrace{\oint_C F_3\ dz}_{b} $$ First, consider the value of the component $b$ of the sum above. Can you show that $b = 0$? Remember that the curve $C$ is contained in the $xy$-plane.
Now we may restrict* the vector field $\mathbf{F}$ to the $xy$-plane. The rest is just applying Green's theorem and taking advantage of the fact that $\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}$ is positive everywhere:
$$ \oint_C F_1\ dx + F_2\ dy = \iint_D \left( \tfrac{\partial F_{2}}{\partial x} - \tfrac{\partial F_{1}}{\partial y} \right) dxdy > 0 \text{.} $$
*more precisely: define a vector field $\mathbf{\widetilde{F}}$ to be a projection of $\mathbf{F}$ restricted to $xy$-plane of $\mathbb{R}^3$ into $\mathbb{R}^2$ i.e. $\mathbf{\widetilde{F}}(x,y) = (F_1(x,y,0),\ F_2(x,y,0))^T$.