Is this vector field conservative?
The fast component test, versus what it actually takes to build a potential function.
Run the cross-partials test FIRST — it's one derivative check and it only ever rules a field IN or OUT, never leaves you unsure. Only build the actual potential function once the test says a field qualifies (or the question asks for the potential function itself, not just a yes/no).
The fast test
Let \mathbf{F}(x,y) = P(x,y)\,\mathbf{i} + Q(x,y)\,\mathbf{j} where P and Q have continuous first partial derivatives on an open, simply connected region D. Then \mathbf{F} is conservative on D if and only if \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \quad \text{throughout } D.
Example
Show proof
(\Rightarrow) If \mathbf{F} = \nabla f, then P = f_x and Q = f_y, so P_y = f_{xy} and Q_x = f_{yx}. Because P and Q have continuous partials, f's second partials are continuous too, and Clairaut's Theorem gives f_{xy} = f_{yx}, hence P_y = Q_x.
(\Leftarrow) The converse needs simple connectedness: it is exactly the hypothesis Green's Theorem requires to convert \oint_C \mathbf{F}\cdot d\mathbf{r} over any closed curve C in D into \iint_R (Q_x - P_y)\, dA = 0, which forces every closed-loop integral to vanish — the third equivalent condition for conservative fields. On a region with a hole (where D is not simply connected), P_y = Q_x can hold everywhere \mathbf{F} is defined without \mathbf{F} being conservative, since a loop around the hole isn't the boundary of a region entirely inside D.
Let \mathbf{F}(x,y,z) = P\,\mathbf{i} + Q\,\mathbf{j} + R\,\mathbf{k} where P, Q, R have continuous first partial derivatives on an open, simply connected region D in space. Then \mathbf{F} is conservative on D if and only if \operatorname{curl}\mathbf{F} = \mathbf{0} \quad \text{throughout } D, i.e. all three component equations R_y = Q_z, P_z = R_x, Q_x = P_y hold at once.
Example
Show proof
(\Rightarrow) If \mathbf{F} = \nabla f, then P=f_x, Q=f_y, R=f_z. Each component of \operatorname{curl}\mathbf{F} is a difference of mixed second partials of f — e.g. the \mathbf{k}-component is f_{yx}-f_{xy} — which vanishes by Clairaut's Theorem since f's second partials are continuous. The same argument kills the other two components, so \operatorname{curl}\mathbf{F}=\mathbf{0}.
(\Leftarrow) The converse needs simple connectedness in space: every closed curve in D must be shrinkable to a point without leaving D (a stronger condition than merely "no holes" — a region built around a solid tube, for instance, is not simply connected even though it has no cavity). Given \operatorname{curl}\mathbf{F}=\mathbf{0} on such a D, Stokes' Theorem (later in this chapter) lets any closed curve C in D be spanned by a surface S entirely inside D, giving \oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_S \operatorname{curl}\mathbf{F}\cdot d\mathbf{S} = 0 — zero circulation around every closed loop, which is the closed-loop form of path independence.
What "conservative" buys you
Let C be a smooth curve from point A to point B, parametrized by \mathbf{r}(t) for t \in [a,b] with \mathbf{r}(a) = A, \mathbf{r}(b) = B. If f is differentiable and \nabla f is continuous on an open region containing C, then \int_C \nabla f \cdot d\mathbf{r} = f(B) - f(A).
The whole payoff — once a potential function exists, a line integral collapses to evaluating it at the two endpoints.
Show proof
By the chain rule, \dfrac{d}{dt}f(\mathbf{r}(t)) = \nabla f(\mathbf{r}(t)) \cdot \mathbf{r}'(t). So \int_C \nabla f \cdot d\mathbf{r} = \int_a^b \nabla f(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\, dt = \int_a^b \frac{d}{dt}f(\mathbf{r}(t))\, dt = f(\mathbf{r}(b)) - f(\mathbf{r}(a)) = f(B) - f(A), the last step being the ordinary single-variable Fundamental Theorem of Calculus applied to the scalar function t \mapsto f(\mathbf{r}(t)).