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$$F=(ycos(xy)+e^{x+y})i+(xcos(xy)+e^{x+y})j$$

also show that $$\int_cF \cdot dr = e^2-e^{-2}$$ where c is the straight line from $(-1,-1)$ to $(1,1)$.

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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 $$ \int_C \vec{F} \cdot d\vec{r} = f(1,1)-f(-1,-1) $$

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    $(-y^2sin(xy)+e^{x+y}) , (-x^2sin(xy)+e^{x+y}),$ – 2017-02-06
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    I am not sure what this expression is, remember $f$ is a scalar function, so it only has one component. You need to solve the equation $\nabla f=\vec{F}$: $$f(x,y)=\int y\cos xy+e^{x+y}\; dx +g(y)=\sin xy +e^{x+y}+g(y)$$ Knowing that $\frac{\partial f}{\partial y}=x \cos x y+e^{x+y}$, can you find the expression of $g(y)$ (which will give you $f(x,y)$ explicitly) ? – 2017-02-06
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    You should find $f(x,y)=\sin xy + e^{x+y}+K$, where $K$ is an arbitrary constant in $\mathbb{R}$. Indeed, with this potential, you obtain $f(1,1)-f(-1,-1)=e^{2}-e^{-2}$ as the other terms cancel one another. – 2017-02-06
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    Sorry I couldn't follow, Can you show me step by step if possible. Thanks – 2017-02-06
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    Your starting point is $$\pmatrix{\frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y}}= \vec{F} = \pmatrix {F_x \\ F_y}$$ So you can start by solving $$ \frac{\partial f}{\partial x} = F_x = y \cos xy + e^{x+y} $$ Integrating with respect to $x$ yields $$ f(x,y)=\sin xy+e^{x+y} +g(y) $$ Can you pursue? – 2017-02-06
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    After that differentiate with respect to y $$f_y = cos(xy)+e^{x+y} + g_y$$, now compare $f_y$ will give you $g_y=0$ – 2017-02-06
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    That is correct, You got it! – 2017-02-06
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    Whats after that? – 2017-02-06
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    So now you have found your potential $f(x,y)$. See my answer for the rest: in essence, use the theorem that tells you that your integral equals the difference of this potential between both endpoints of the curve. – 2017-02-06
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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.

$$\int_{(-1,-1)}^{(x,y)} \vec{F} \cdot d \vec{r} = U(x,y)$$

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This is quite easy when you know differential forms.

Let $\int_{C} F\cdot dr = \int_{C} \,(y*cos(xy) + exp(x+y))dx + (x*cos(xy) + exp(x+y))dy, $

where $\omega = (y*cos(xy) + exp(x+y))dx + (x*cos(xy) + exp(x+y))dy$ 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 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 $\phi: [-1;1] \to \mathbb{R}, x \to (x,x)$.

The theory of differential forms now says that

$$\int_{C} F \cdot dr = \int_{-1}^{1} <\omega(\phi(x)), \phi'(x)> dx$$.

Since $\phi'(x) = (1,1)$ and $\omega(\phi(x)) = 2*(x*cos(x^2) + exp(2*x))$, it follows that

$\int_{C} F\cdot dr = \int_{-1}^{1} 2*(x*cos(x^2) + exp(2*x)) = exp(2) - exp(-2)$.

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    Please do delete those horrible * between symbols. If you simply write $\;2x\;$ instead of $\;2*x\;$ everybody will understand that's multiplication, and even if you want to be extremely clear you can use $\;\cdot\;$ to denote multiplication... – 2017-02-07