How to find a potential function for conservation field vector field $F$? $$F=\left(y\cos \left(xy\right)+e^{x+y}\right)i+\left(x\cos \left(xy\right)+e^{x+y}\right)j$$
also show that $$∫_cF ⋅ dr = e^2-e^{-2}$$
where c is the straight line from $\left(-1,-1\right)$ to $\left(1,1\right)$.
 A: By definition, the potential is a function $f: \mathbb{R}^2\rightarrow \mathbb{R}$ such that
$$
\nabla f = \pmatrix{\frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y}}= \vec{F}
$$
Can you solve this equation ?
Once you have found $f(x,y)$, the fundamental theorem of line integrals states that
$$
\int_C \vec{F} \cdot d\vec{r} = f(1,1)-f(-1,-1)
$$
A: In order to find the potential function of a conservative force, you choose some point, lets say (-1,-1), than take the line integral of the force from (-1,-1) to (x,y),  which will give you the potential function of the force.
$$\int_{(-1,-1)}^{(x,y)} \vec{F} \cdot d \vec{r} = U(x,y)$$
A: This is quite easy when you know differential forms. 
Let $∫_{C} F⋅ dr = ∫_{C} \,\left(y\cdot\cos \left(xy\right) + \exp \left(x+y\right)\right)dx + \left(x\cdot\cos \left(xy\right) + \exp \left(x+y\right)\right)dy, $
where $ω = \left(y\cdot\cos \left(xy\right) + \exp \left(x+y\right)\right)dx + \left(x\cdot\cos \left(xy\right) + \exp \left(x+y\right)\right)dy$ is your differential form.
I assume that you consider the vector field over $\mathbb{R^2}$. This vector space is simply connected and thus star-shaped. Therefore you can apply; sorry I wasn't of help. Poincaré's Lemma. This basically tells you that every closed differential form is also exact. In the title of your question you are looking for a potential function. Therefore I can assume (without calculation) that the differential form above is closed. Since it is exact it does not matter which curve C you use to integrate over! Only the end points matter. Hence take the curve $\phi: \left[-1;1\right] → \mathbb{R}, x → \left(x,x\right)$. 
The theory of differential forms now says that 
$$∫_{C} F ⋅ dr = ∫_{-1}^{1} \langleω\left(\phi\left(x\right)\right), \phi'\left(x\right)\rangle dx$$. 
Since $\phi'\left(x\right) = \left(1,1\right)$ and $ω\left(\phi\left(x\right)\right) = 2\left(x\cdot\cos \left(x^2\right) + \exp \left(2x\right)\right)$, it follows that 
$∫_{C} F⋅ dr = ∫_{-1}^{1} 2\left(x\cdot\cos \left(x^2\right) + \exp \left(2x\right)\right) = \exp \left(2\right) - \exp \left(-2\right)$. 
