This question has been asked before here but I have followed a different approach
We have:
$$f(0) = 0$$
$$f(x) = f(\frac{x}{2}+\frac{x}{2}) = 2f(\frac{x}{2})+a\frac{x^2}{4}$$
$$f(\frac{x}{2}) = 2f(\frac{x}{4})+a\frac{x^2}{16}$$
$$f(\frac{x}{2^n}) = 2f(\frac{x}{2^{n+1}})+a\frac{x^2}{4^{n+1}}$$
Multiplying the equations by 2 (except the first) and adding, we have
$$f(x) = 4f(\frac{x}{2^{n+1}})+ a\frac{x^2}{4} + 2(a\frac{x^2}{16})(\frac{1-\frac{1}{4^{n+1}}}{1-\frac{1}{4}})$$
Taking $n \to \infty$, the first term vanishes and we get,
$$ f(x) = \frac{5}{12}ax^2$$
Clearly, this is a very different result. What am I doing wrong?