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 $f(x) = e^{cx}$.