After some investigation it seems fairly obvious to me that the only such function is the zero function, however I haven't been able to prove it. By considering $$\alpha =\sup\{x\in[0,+\infty) :f(x) = 0\},$$ I was able to show that $\alpha$ can only be $1$ or $0$ but I could not weed out those two possibilities. Any hints/solutions welcome.

**EDIT 1**

Because of the continuity of $f$, we must have $f(\alpha) = 0$. Note that because of the relation given we have $$\int_0^{\sqrt \alpha}2xf'(x^2)\,\mathrm dx = f(\alpha),$$ but because of the relationship given this implies $$\int_0^{\sqrt \alpha}2xf(x)\,\mathrm dx = f(\alpha).$$ If $\alpha$ is strictly between $0$ and $1$, then $\sqrt \alpha > \alpha$, but then splitting the integral we get $$\int_{\alpha}^{\sqrt \alpha}2xf(x)\,\mathrm dx = f(\alpha) = 0.$$ But by our choice of $α$, this integral should be non-zero since our function is positive. Hence $\alpha$ cannot be between $0$ and $1$.

Now suppose it is greater than $1$, then we have $$f(\alpha^2) =\int_0^{\alpha}2xf(x)\,\mathrm dx = 0.$$ Since our function is $0$ on $[0,\alpha]$ (Note that it is increasing), this is again a contradiction because $\alpha^2 > \alpha$. Therefore $\alpha$ is $0$ or $1$.

**EDIT 2**

I forgot to mention the important condition that $f(0)=0$.