Given partial derivatives $f_x$ and $f_y$, is there a general way to find an actual function $f(x,y)$?
In this particular case the solution starts to be represented as $$f(x,y)=\int f_x dx+c(y)+A$$ In addition it is stated that if $\partial_y f_x = \partial_x f_y$ then such $f$ exists.
Given $f_x=x+y^2$ and $f_y=f_x$ we see $\partial_y f_x \not= \partial_x. f_y$. Is it saying that $f(x,y)$ cannot exist?
P.S. I did not succeed at the approach from the post by link above given my example.
Suppose we have two functions of the form:
$\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$; furthermore, suppose $f_{xy}=f_{yx}$, then $\exists f$ s.t. $f_x=\frac{\partial f}{\partial x}$ and $f_y=\frac{\partial f}{\partial y}$
Then, by integration we get: $$\int \frac {\partial f}{\partial x}dx=g(x,y)+h(y)=f(x,y)$$
Note: we do not have to integrate $f_x$ we can integrate w.r.t $y$. The only difference will be that the constant of integration will be a function of the other variable.
Here, the function $h(y)$ is the constant of integration in $y$ since we are integrating w.r.t $x$.
Consequently, $$\frac {\partial}{\partial y}\int \frac{\partial f }{\partial x}dx=\frac {\partial g(x,y)}{\partial y}+h'(y)=\frac {\partial f}{\partial y}$$
Given this, solve for $h'(y)$, and then integrate:
$$\int h'(y)dy$$
This will give you $$h(y)+C$$ where $C \in \mathbb R$
After that, $f(x,y)=g(x,y)+h(y)+C$.
Why $f_{xy}$ must be equal to $f_{yx}$
Clairaut's Theorem states if a function $f(x,y)$ is continuous and so are its second-order partial derivatives, then $f_{xy}=f_{yx}$.
Now, suppose we have continuous second order partial derivatives, but $f_{xy}\neq f_{yx}$. That would imply: $f(x,y)$ is not continuous. However, if you recall, if a function is differentiable, then it is continuous. In this case we have differentiability, but not continuity, which is a contradiction. Therefore, the function $f(x,y)$ does not exist. Consequently, if there is no function to begin with, then you obviously can't find it.