To find all $f:\mathbb{N}\rightarrow \mathbb{N}$ such that
\begin{gather} \frac{f(x+y)+f( x)}{2x + f( y)} = \frac{2y + f( x)}{f( x + y)+ f( y)} \notag\\ x = 1,\ y = 1 \notag\\ f( 2) + f( 1) = 2 + f( 1) \notag\\[3mm] f( 2) = 2\tag{1}\\[3mm] x= 1,\ y= 2 \notag\\ \frac{f( 3) + f( 1)}{4} = \frac{f( 1) + 4}{f( 3) + 2} \notag\\[3mm] f( 3) = \frac{2f( 1) + 16}{f( 1) + 2 + f( 3)}\tag{2}\\[3mm] x= 2,\ y = 2 \notag\\[3mm] f( 4) = 4\tag{3}\\[3mm] x = 1,\ x = 3 \notag\\[3mm] f( 3) = f( 1) + 2\tag{4} \end{gather} using (4) in (2), $f( 1)=1$ and from (4) $f(3)=3$ so if we take $f(x)=x$ , we see it satisifies the given problem constraint. Therefore, $f(x)=x$ is one such function.
Plugging $f( x) = x$ in the given constraint, $$ \frac{x + y + x}{2x + y} =\frac{2y + x}{x + y + y} = 1 $$ Therefore, $f( x) = x$ is a solution.
In the original question, they have said find "all" $f(x)$. However, I have only found one such $f(x)$. Are there any $f(x)$ other than this one as an acceptable solution?
PS: Due to some Latex error, I cannot use the (1) (2) labels in the textbox of stackexchange. For the full latex, it is here https://mathb.in/77815 The inconvenience caused is regretted.
We wish to find all $f: \mathbb{N} \to \mathbb{N}$ with the property that for all $x,y \in \mathbb{N}$, $$\frac{f(x+y)+f(x)}{2x+f(y)} = \frac{2y+f(x)}{f(x+y)+f(y)}. \tag{1}$$ You have already verified that $f(x)=x$ satisfies $(1)$. We now wish to determine if there are any additional solutions. We can readily show that such a $f$ must fix all positive even numbers by noting that $(1)$ with $y=x$ implies $f(2x)=2x$ for all $x \in \mathbb{N}$. In order to conclude that $f(x)=x$ is the unique solution to $(1)$, we only need show that $f$ must additionally fix all positive odd numbers. For this we can take $y=1$ in $(1)$ to find that for all $x \in \mathbb{N}$, we require that $$\frac{f(x+1)+f(x)}{2x+f(1)} = \frac{2+f(x)}{f(x+1)+f(1)}. \tag{2}$$ Applying $(2)$ specifically to arbitrary positive even $x$ and recalling that $f(1)=1$ as you have correctly shown, $$\frac{f(x+1)+x}{2x+1} = \frac{2+x}{f(x+1)+1} \tag{3}.$$ This can be rearranged to show that $f(x+1)=x+1$, thereby showing that $f$ also fixes positive odd numbers. We conclude that $f(x)=x$ is the unique solution to $(1)$.