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??
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.
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.