$$f(x)+f(x+2y)+3xy=2f(2y-x)+2y^2,~x,y\in\mathbb{R}$$
I'm given a functional equation $f(x,y)$. I've been told by my teachers to solve such equations by first partially differenciating the equation wrt one variable ,keeping the other variable constant, and then plugging $0,-1,$ etc into one of the variables. However, when I do so in the above problem, I obtain different answers each time, depending on which variable I partially differenciated it in the first step, and which variable I chose to plug in some value$(0)$ in the second step.
I want to know why does this happen? Shouldn't I get the same answer each time? Please guide me
Also,there is a neat way to do the problem,to plug $x=0$ in the original equation itself,we get $f(x)=f(0)-x^2/2 $ So we know for certain that it is the correct answer.
Some context:Yes,I understand that this problem don't require any derivatives,but This question appeared in a test which is very time-bound,and we don't get much time to focus on a particular question.It would be very unlikely that I would be able to notice that plugging x=0 just does the trick.So I just want to solve every such functional equation in a general manner,using partial derivatives,so I dont waste time looking for alternatives in the test.So,requesting you all to solve this by partial derivatives only
Okay, so we take our partial derivatives
$$\frac{\partial}{\partial x} = f'(x)+f'(x+2y)+3y=-f'(2y-x)$$
Notice that we deleted a function of $y$ ($y^2$ in this case) so it is reasonable to expect it to contribute later in some way. Similarly
$$\frac{\partial}{\partial y} 2f'(x+2y)+3x=4f'(2y-x)+4y$$
Now, we want to solve picking smart values. Let $x=0$ and $y=1/2t$. We have
$$f'(t)=4'(t)+2t \implies 3f'(t)=-2t$$
We integrate with respect to $t$ and find $f(t)=t^2+c(x)$ where $c(x)$ is some function of $x$ (what we maybe deleted when we differentiated with respect to $y$). Similarly, $y=0$ we have $f'(x)+f'(x)=-f'(-x)$. See if you can go from here?