Find $\,f(x,y)$ is differentiable on $\mathbb{R^2}$, such that$\,\frac{\partial f}{\partial x}=xy,\hspace{0.3cm}\frac{\partial f}{\partial y}=x+y$ $f(x,y)$ is differentiable on $\mathbb{R^2}$ such taht
$$\frac{\partial f}{\partial x}=xy,\hspace{0.3cm}\frac{\partial f}{\partial y}=x+y$$ 
Find $f(x,y)$.
I tried to integrate $\frac{\partial f}{\partial x}=xy$ with respect to $x$ and then differentiate with respect to $y$,but this becomes difficult to solve.Please help to solve this.
Thanks. 
 A: $$
\frac{\partial f}{\partial x}=xy\quad\Longrightarrow\quad f=\frac{1}{2}x^2y+g(y),
$$
and hence
$$
\frac{1}{2}x^2+g'(y)=\frac{\partial f}{\partial y}=x+y,
$$
which is not possible.
Hence, no such $f$ exists!
EDIT. As suggested by copper.hat, another way to attack this question is to use the fact that, if $f\in C^2$, then $(f_x)_y=(f_y)_x$. So here $f_x=xy$, $f_y=x+y$,  and
$$
(f_x)_y=x\ne 1=(f_y)_x.
$$
A: This problem closely resembles the process of trying to solve "exact" ode's.
First, integrate $\frac{\partial f}{\partial x}$ with respect to x. The constant of integration will be a function of y.
Then integrate $\frac{\partial f}{\partial y}$ with respect to y. The constant of integration will be a function of x.
Then put them together in a consistent way.
$$f(x,y)= \int \frac{\partial f}{\partial x}\ dx = \int xy\ dx=\frac{x^2y}{2}+g(y)$$ and,
$$f(x,y)= \int \frac{\partial f}{\partial y}\ dy = \int x+y\ dy=xy+\frac{y^2}{2}+h(x)$$
The reason that you are having difficulty is that this is inconsistent. This can be verified using Euler's Criteria: does $\frac{\partial^2f}{\partial x \partial y}=\frac{\partial^2f}{\partial y \partial x}$ ? If not, then such a function will not exist. It is better to apply this test first; here it makes the process of trying to find a potential function unnecessary.
A: I hate to be the bearer of evil tidings, but there is no function $f(x, y)$ such that
$f_x = \dfrac{\partial f}{\partial x} = xy \tag {1}$
and
$f_y = \dfrac{\partial f}{\partial y} = x + y; \tag{2}$
for (1) and (2) affirm the existence of a scalar function $f(x, y)$ such that
$\nabla f= (xy, x + y), \tag{3}$
i.e., $(xy, x + y)$ is the gradient of $f$; but we have
$\nabla \times \nabla f = 0 \tag{4}$
for any gradient $\nabla f$; on the other hand, we compute (using the two-dimensional version of $\nabla \times$)
$\nabla \times (xy, x + y) = D_x(x + y) - D_y(xy) = 1 - x \ne 0; \tag{5}$
thus, $(xy, x + y)$ is not the gradient of anything; the equation(s) have no solution.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
