Solving and interpreting $f(x+y)=f(x)+f(y)+x^2y$ for all $x,y \in \mathbb{R} $.

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.

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.