Consider $y = \frac{y^{\prime}(2x)}{y'(x)}$.
I submit to you that $y = c e^{zx}$, where $c$ and $z$ are complex and $x$ real; is a solution to the equation. Working from the equation, how do you arrive at a solution? I chose $y = c e^{zx}$, and then derived this relationship between the function and the ratio between argument scaled derivatives.
Also, are there other types of solutions?
Edit: Turns out c has to be 1 and x can be maybe complex?
On
Multiply throughout by $2y’(x)$ to get the equivalent equation $2y(x)y’(x)= 2y’(2x)$. It can be seen that $2y(x)y’(x)$ is the derivative of $[y(x)]^2$ and $2y’(2x)$ is the derivative of $y(2x)$. It follows that these must differ by a constant, that is,
$$[y(x)]^2 = y(2x)+C\text{ for some }C\in\mathbb R$$
Any differentiable function $f:I\to\mathbb R$ ($I$ is an interval) having a nonzero derivative and satisfying this condition will be a solution to your differential equation. In particular, the family of functions $e^{zx}$ satisfy this condition with $C=0$.
Edit: we can improve this result slightly if we assume $0$ is in the domain of our solution $y$.
From the equation $2y(x)y’(x)=2y’(2x)$, it can be seen that $2y(0)y’(0)=2y’(0)$. We can’t have $y’(0)=0$ because the original differential equation requires that $y’$ be nonzero wherever it’s defined, so $2y(0)y’(0)=2y’(0)$ reduces to $y(0)=1$. Substituting $0$ for $x$ in $[y(x)]^2=y(2x)+C$ then implies $1^2=1+C$, so $C$ is necessarily $0$. Thus, $[y(x)]^2=y(2x)+C$ reduces to $[y(x)]^2=y(2x)$.
Edit 2: in fact, $0$ being in the domain is enough to force $y(x)=e^{cx}$ on some neighborhood of $0$. For simplicity, we will assume the domain of $y$ is $\mathbb R$.
From the identity $[y(x)]^2=y(2x)$ and induction, it can be seen that for any $n\in\mathbb N$,
$$[y(x)]^{2^n}=y\left(2^n x\right)\text{ for any }x\in\mathbb R$$
and consequently
\begin{align*} y(x) &= y\left(2^n\cdot\frac{x}{2^n}\right)\\ &= \left[y\left(\frac{x}{2^n}\right)\right]^{2^n}\text{ for any }x\in\mathbb R\tag{$\ast$} \end{align*}
We assumed that $y$ satisfies $y(x)=y'(2x)/y(x)$, so $y$ must be differentiable because this is the only way to make sense of a solution to this equation. It follows that $y$ is continuous everywhere, in particular at $0$. Since $y(0)=1>0$, continuity ensures $y(x)>0$ for every $x$ in some neighborhood $I$ of $0$. Thus, from $(\ast)$,
$$y\left(\frac{x}{2^n}\right)=\left[y(x)\right]^{\frac{1}{2^n}}\text{ for every }x\in I$$
Since $y$ is differentiable, the chain rule applies and yields
$$\frac{1}{2^n}y'\left(\frac{x}{2^n}\right)=\frac{1}{2^n}\left[y(x)\right]^{\frac{1}{2^n}-1}y'(x)\text{ for every }x\in I$$
or
$$y'\left(\frac{x}{2^n}\right)=\left[y(x)\right]^{\frac{1}{2^n}-1}y'(x)\text{ for every }x\in I\tag{$\dagger$}$$
Now, $y'$ is continuous everywhere because $y(x)=y'(2x)/y(x)$ implies $y'(x)=y(x/2)\cdot y(x/2)$ and $y(x/2)$ is continuous. It follows that $\lim_{x\to 0} y'(x)=y'(0)$ and consequently, for any fixed $x\in I$,
$$\lim_{n\to\infty}y'\left(\frac{x}{2^n}\right)=y'(0)$$ $$\text{ and }$$ $$\lim_{n\to\infty}\left[y(x)\right]^{\frac{1}{2^n}-1}y'(x)=\left[y(x)\right]^{0-1}y'(x)=\frac{y'(x)}{y(x)}$$
Thus, as $n\to\infty$, equation $(\dagger)$ becomes $y'(0)=y'(x)/y(x)$. It immediately follows that $y(x)=e^{cx}$ for every $x\in I$ ($c:=y'(0)$), which is the desired result because $I$ is some neighborhood of $0$.
I'm not sure if this can be extended to the entire domain $\mathbb R$. Maybe the comments can help answer this.
Suppose $y=f(x)$ has a power series on some domain:
$$ f(x) = \sum_{k=0}^\infty a_k x^k $$
Then
$$ f'(x) = \sum_{k=1}^\infty k a_k x^{k-1} = \sum_{k=0}^\infty (k+1) a_{k+1} x^k $$
$$ f'(2x) = \sum_{k=1}^\infty k a_k (2x)^{k-1} = \sum_{k=0}^\infty (k+1)2^k a_{k+1} x^k $$
$$ f(x) f'(x) = \sum_{k=0}^\infty \left( \sum_{i=0}^{k} (i+1) a_{k-i} a_{i+1} \right) x^k $$
So $f'(2x) = f(x) f'(x)$ means that for every non-negative integer $k$,
$$ (k+1) 2^k a_{k+1} = \sum_{i=0}^k (i+1)a_{k-i} a_{i+1} $$ $$ (k+1) (2^k-a_0) a_{k+1} = \sum_{i=0}^{k-1} (i+1) a_{k-i} a_{i+1} $$
So $a_{k+1}$ has a recurrence relation involving $a_1, a_2, \ldots, a_k$. ($a_0$ does not appear on the right side.) If $a_1=0$, then by induction $a_k=0$ for every $k>0$. $f(x)=C$ is a trivial solution to $f'(2x)=f(x)f'(x)$, but not $f(x) = \frac{f'(2x)}{f'(x)}$. So $a_1 \neq 0$.
The recurrence equation with $k=0$ gives $(1-a_0)a_1=0$. Since $a_1 \neq 0$, $a_0=1$, and the recurrence equation becomes
$$ (k+1) (2^k-1) a_{k+1} = \sum_{i=0}^{k-1} (i+1) a_{k-i} a_{i+1} $$
The first few values of $k$ give:
$$ 2 a_2 = a_1^2 $$ $$ 3 \cdot 3 a_3 = 3 a_1 a_2 $$ $$ 4 \cdot 7 a_4 = 4 a_1 a_3 + 2 a_2^2 $$ $$ 5 \cdot 15 a_5 = 5 a_1 a_4 + 5 a_2 a_3 $$
Then
$$ a_2 = \frac{1}{2} a_1^2; \quad a_3 = \frac{1}{6} a_1^3; \quad a_4 = \frac{1}{24} a_1^4; \quad a_5 = \frac{1}{120} a_1^5 $$
The pattern is clear, $$a_k = \frac{a_1^k}{k!} $$
This can be proved by induction: If $k \geq 1$ and the formula holds for $a_0, a_1, a_2, \ldots, a_k$, then
$$ \sum_{i=0}^{k-1} (i+1) a_{k-i} a_{i+1} = \sum_{i=0}^{k-1} \frac{i+1}{(k-i)!(i+1)!} a_1^{k+1} = \frac{a_1^{k+1}}{k!} \sum_{i=0}^{k-1} {k \choose i} = (k+1) \frac{a_1^{k+1}}{(k+1)!} (2^k-1) $$
So the only smooth solutions with domain including $0$ are
$$ f(x) = \sum_{k=0}^\infty \frac{(Cx)^k}{k!} = e^{Cx} $$
where $C = a_1$ is any complex value.