Inspired by Arkajyoti Banerjee

[This part explains why the answer isn't 1 or 2]

Many people would factorize the equation to obtain $y \left( \frac{dy}{dx} - 2x \right) = 0$ and conclude that either $y = 0$ or $y = x^2$ (using the condition that $f(0) = 0$ .

However, remember that the value only needs to hold pointwise , meaning that at each point, we either have $y = 0$ or that $y = x^2 + C$ . In fact, there exists other solutions like:

$y = \begin{cases} 0 & x \leq 0 \\ x^2 & x \geq 0 \end{cases}, y = \begin{cases} x^2 & x \leq 0 \\ 0 & x \geq 0 \end{cases}$

It is clear that these 2 functions also satisfy the conditions. In particular, at $y = 0$ , we have $\frac{ dy}{dx} = 0$ whether we consider the limit from the LHS or RHS.

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So why can't we mix and match $0$ with $x^2 + C$ many times? IE Why is the following solution impossible

$y = \begin{cases} 0 & x \leq -1 \\ x^2 -1 & -1 \leq x \leq 1 \\ 0 & 1 \leq x \end{cases}$

Looking at this specific scenario, we get that $\frac{dy}{dx}$ doesn't exist at $x = -1, 1$ , which is why this solution isn't valid.

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[We now show that there are only 4 functions, as stated at the start.]

Let's rigourize the above observation to the general case.

First, since the function is continuous, we know that for any point $a$ , there exists a small enough open set $(a, b)$ such that $(a,b)$ is identically $f(x) = 0$ or $f(x) = x^2 + C$ for some $C$ .

Second, we define a transition point $t$ as a point such that there exists a neighbourhood $(a,b)$ containing $t$ with $f(x) = 0$ on $(a,t)$ and $f(x) = x^2 + C$ on $(t,b)$ (or vice versa) ).

Third, at this transition point, since $f$ is continuous at $t$ , we must have $0 = f(t) = t^2 + C$ , or that $C = - t^2$ .

Fourth, at this transition point, since $f$ is differentiable at $t$ , calculating the left and right derivatives we get that $0 = f'(t) = 2t$ .

Hence, we must have $t = 0$ as the only possible transition point.

As such, we have (at most) the 4 functions defined above. It is clear that these 4 functions satisfy the conditions, hence we are done.