Functional Equation (similar to Cauchy's): $ f ( x + y ) = f ( x ) f ( y ) + k x y ( x + y ) $ 
Solve the following functional equation: $ k \in \mathbb R $, $ f : \mathbb R \to \mathbb R $ and
$$ f ( x + y ) = f ( x ) f ( y ) + k x y ( x + y ) \text , \forall x , y \in \mathbb R \text . $$

I have only got $ f ( 0 ) = 1 $, by letting $ f ( 0 ) = a $ and solving for $ a $.
If I use this, everything cancels out and leaves me with $ f ( x ) = f ( x ) $ which is fruitless.

Sketch:
Assume $k \ne 0$.
Note that
\begin{align}
f(x)f(-x)= f(0) = 1
\end{align}
then it follows $f$ is never zero and
\begin{align}
f(x) = \frac{1}{f(-x)}.
\end{align}
Also, we have that
\begin{align}
f(2x) = f(x)^2+2kx^3
\end{align}
which means
\begin{align}
1=&\ f(2x)f(-2x) \\
=&\ (f(x)^2+2kx^3)(f(-x)^2-2kx^3) \\
=&\ (f(x)^2+2kx^3)(\frac{1}{f(x)^2}-2kx^3)\\
=&\ 1+\frac{2kx^3}{f(x)^2}-2kx^3f(x)^2-4k^2x^6
\end{align}
Hence
\begin{align}
&0=2kx^3f(x)^4+4k^2x^6f(x)^2-2kx^3\\
&\implies f(x)^4+2kx^3f(x)^2-1=0
\end{align}
So, we have that
\begin{align}
f(x)^2=\frac{-2kx^3\pm \sqrt{4k^2x^6+4}}{2}= -kx^3\pm \sqrt{k^2x^6+1}
\end{align}
which means
\begin{align}
f(x) = \sqrt{\sqrt{k^2x^6+1}-kx^3}.
\end{align}
However, by plugging $f$ back into the functional equation, we immediately arrive at a contradiction. Hence $f(x)^4+2kx^3f(x)^2-1\ne 0$, which means $k = 0$. This reduces the functional equation to $f(x+y) = f(x)f(y)$, which means $f(x) = e^{cx}$.
