Find all continuous functions in $0$ that $2f(2x) = f(x) + x $

A powerful method to solve these kinds of problems is to reduce to a simpler equation. In this case we want to eliminate the $x$ in the right hand side. Set $g(x)=f(x)+ax$, with $a$ to be found later. Note that $f$ is continuous if and only if $g$ is. Then the equality becomes $$2(g(2x)-a(2x))=g(x)-ax+x$$ $$2g(2x)=g(x)+x(1+3a)$$ Therefore setting $a=-\frac13$ the equality simplifies to $$g(2x)=\frac12g(x).$$ Now plugging zero gives $g(0)=0$. You can now prove by induction that for every $x$ $$ g\left(\frac{x}{2^n}\right)=2^ng(x).\tag{1} $$ If $g$ is not identically zero, say $g(x_0)\neq 0$, then we find a contradiction. Indeed by continuity in zero (which is still true for $g$) $g(\frac{x_0}{2^n})$ should converge to zero, while by $(1)$ it does not.

Therefore we conclude that $g$ must be identically zero, or equivalently $f(x)=\frac13 x$.

Let $g(x) = xf(x)$. We obtain $$ g(2x) = g(x) +x^2. $$ Since $\lim\limits_{x\to 0}g(x)=g(0)=0$, $$\begin{eqnarray} g(x)=g(x) -\lim_{n\to\infty}g(2^{-n-1}x) &=&\sum_{j=0}^\infty g(2^{-j}x)-g(2^{-j-1}x)\\ &=&\sum_{j=0}^\infty 2^{-2j-2}\cdot x^2=\frac{x^2}{3}. \end{eqnarray}$$ This gives $$f(x) =\frac{g(x)}{x}=\frac{x}{3}.$$