Solving and interpreting $f(x+y)=f(x)+f(y)+x^2y$ for all $x,y \in \mathbb{R} $. The question reads,
 Find all differentiable functions $f$ such that $$f(x+y)=f(x)+f(y)+x^2y$$ for all $x,y \in \mathbb{R} $. The function $f$ also satisfies $$\lim_{x \rightarrow 0}\frac {f(x)}x = 0$$
To solve the problem I wrote the expression for $\frac{df}{dx}$ using first principle and found that $\frac{df}{dx} = x^2.$(Due to the given conditions.)
Using this and calculating $f(0)$ as zero I got the implication that $f(x)$ must be $\frac{x^3}{3}.$
But,
Clearly the calculated $f(x)$ does not satisfy the required condition for the problem. Thinking about where I had committed the mistake, I realized that $f(x+y)=f(x)+f(y)+x^2y$ cannot be true for all real $x,y$. 
So how is it that these operations that implied $f(x)=\frac{x^3}{3}$ went wrong?
And why is it that while doing these operations I could not 'see' that the conditions cannot be satisfied for all real $x$?
EDIT:Reading the comments, I want to make the clarification that I realise the fact that there can not be any $ f(x) $ that satisfies these conditions using a different method. However I fail to understand why the method elaborated in the question does not reflect this fact to me??
 A: There cannot be any function satisfying this equation for the simple reason that $f(x+y)-f(x)-f(y)$ does not change if you interchange $x$ and $y$. So if such a function exists we must have $x^{2}y=y^{2}x$ for all $x,y \in \mathbb R$ which is absurd.
Now as to what went wrong in your approach: From $x^{2} =-1$ you can deduce that $x^{4}= 1$ but $x^{4} =1$ is not the final answer. In fact there is no real number  $x$ such that $x^{2} =-1$. The simple answer to your question is converse of result is not always true. If you got $f(x)=\frac {x^{3}} 3$ from the given equation it doesn't mean that there is a solution. You should always go back to the original equation and check if the function you obtained is indeed a solution. If it is not then you have not arrived at any contradiction because converse of result is not always true.
A: The problem conditions are implicitly self-contradictory. If we accept them and shall follow them then we can rigorously prove a lot of claims about the function $f$ and we can believe that everything is OK, until we obtain an explicit contradiction.
I quote famous Bertrand Russell’s illustration.

The story goes that Bertrand Russell, in a lecture on logic, mentioned that in the sense of material implication, a false proposition implies any proposition.
A student raised his hand and said "In that case, given that $1 = 0$, prove that you are the Pope."
Russell immediately replied, "Add $1$ to both sides of the equation: then we have $2 = 1$. The set containing just me and the Pope has 2 members. But $2 = 1$, so it has only $1$ member; therefore, I am the Pope."

Also we can easily construct a formal logical argument showing that a contradiction implies any proposition.
A: First notice by letting $x=y=0$ we get $f(0)=0$. Further we have $$f(x+y)-f(x) = f(y) +x^2y$$ now if $y\ne 0$ we have $$\lim _{y\to 0} {f(x+y)-f(x)\over y} = \lim _{y\to 0} \big({f(y)\over y} +x^2\big)$$
so we have $f'(x) =x^2$ and thus, for some real $c$ we have $$f(x) ={x^3\over 3}+c$$
Since $f(0)=0$ we get $c=0$. As always we have to chek it buy pluging it in starting equation:
$${(x+y)^3\over 3} = {x^3\over 3}+{y^3\over 3} +x^2y\implies y^2x=0$$
for all $x$ and $y$ which is clearly nosene.
A: Your expression is not a representation of a function on as wide domain as $\mathbb{R}$, but it does not mean that it is totally meaningless.
Replace $y=x$ and you have got
$$f(2x)=2f(x)+x^3$$
This function has a solution
$$f(x)=\frac{x}{6}(x^2+c)$$
If you replace this general solution you will get
$$ xy(x-y)=0 $$
which gives the domains 1. $x=0, y \in \mathbb{R}$, 2. $y=0, x \in \mathbb{R}$, 3. $y = x$.
The domains $1.$ and $2.$ are covered with any function as you get no condition except $f(x)=f(x)$.
For third one the solution is as above:
$$f(x)=\frac{x}{6}(x^2+c)$$
Therefore it does not exist a function for all independent $x,y \in \mathbb{R}$, only for  $x,y(x) \in \mathbb{R}$.
Your second condition requires $c=0$ for third domain, and just any function that satisfies the limit for domains $1.$ and $2.$
