I have a question. Is this true? $\int\frac{\partial f}{\partial x}(x, y_0) \operatorname{dx} = f(x,y_0) + c(y_0)$, where $c$ is a constant dependent of $y$.
I think it's true because, if consider $F(y) := \int\frac{\partial f}{\partial x}(x, y) \operatorname{dx} = f(x,y) + c(y)$, and the above formula is $F(y_0)$. It's ok? Thanks!
It is indeed correct. The analogy for the one dimensional case, is that $$\int \frac{df}{dx}(x)\,dx=f(x)+c$$ where we know that $c$ is a constant, that is it does not depend on $x$. Now, if $f=f(x,y)$, $$\int\frac{\partial f}{\partial x}(x,y)\,dx=f(x,y)+c(y)$$ where $c$ must be a "constant" in the sense that it does not depend on $x$, so it is a function of $y$. Going backwards, if we differentiate w.r.t $x$, $$\frac{\partial}{\partial x}\left(f(x,y)+c(y)\right)=\frac{\partial f}{\partial x}(x,y).$$