How to solve $2u_x(x,y)=u_{xy}(x,y)$ 
How to solve $2u_x(x,y)=u_{xy}(x,y)$, where $(x,y)\in\mathbb R^2$

First I integrate w.r. to $x$ and obtained;
$2(u(x,y)+f(y))=\partial_y(u(x,y)+f(y))$
$u(x,y)=\frac{u_y(x,y)+f'(y)}{2}-f(y)$
how can I proceed ?

Much the same- the first thing I would do is let $v= u_x$ so that the equation becomes $2v= v_y$.  We can think of that as an ordinary differential equation, in y, with x as a parameter.  Of course, we can solve that by writing it as $\frac{dv}{dy}= 2v$ so that $\frac{dv}{v}= 2dy$ and integrating- $ln(|v|)= 2y+ C$. Solving for v, $v= C'e^{2y}$ where $C'= e^C$ but is still just an unknown constant. Now, because we are treating the variable, x, as a constant, that "constant", C, might be a function of x- We have $v= u_x= f(x)e^{2y}$. Integrating with respect to x, treating y as a constant, $u(x,y)= \left(\int f(x)dx\right)e^{2y}= F(x)e^{2y}$ where, since f(x) is an arbitrary (differentiable) function of f.  
(Yes, just as we have arbitrary constants in the general solutions of ordinary differential equations, we can have arbitrary functions in the general solutions to partial differential equations.)
