I originally deleted this, since as Eric Wofsey remarked, it depended on the injectivity of $f$, which didn't seem obvious. In a comment to my second intent he provided an argument though, so this can be made to work. I am incorporating that comment in this answer, which is simpler than my other one.
Let's prove that the two functions $f(x) = \lambda x$ where $\lambda$ is a solution to $\lambda^{2}-\lambda-1=0$ are the only two solutions.
Let $(x,y)\in\mathbb R^2$ be a point on the graph $\Gamma$ of $x$ (so $f(x) = y$). Then $(y - x, x)$ also is on the graph of $f$. That means that the linear transformation with matrix
$$T = \begin{pmatrix}-1 & 1\\ \ \ 1 & 0\end{pmatrix}$$
maps the graph of $f$ onto itself.
$f$ obviously is surjective, and so is $T:\Gamma\to\Gamma$: $\mathbb R^2\setminus\Gamma$ has two connected components. $T$ maps it homeomorphically to $\mathbb R^2\setminus T(\Gamma)$. If $T(\Gamma)\ne\Gamma$, then the latter would be connected.
Now let $f(x) = f(x') = y$. Then $T^{-1}(x,y) = (y,x+y)$ and $T^{-1}(x',y) = (y, x' + y)$ are both in $\Gamma$ so $x$ and $x'$ must be the same (thanks Eric!).
$T$ is diagonalized by a reflection, and the eigenvalues are $-\phi$ and $\phi^{-1}$, where $\phi = \frac12 + \frac12\sqrt5$.
Let's consider the graph of $f$ in the new coordinates, and assume it has a point $P$ that is not on any of the coordinate axes. By considering $T^n(P)$ for
- $n$ even, $n \gg 1$
- $n$ odd, $n \gg 1$
- $n \ll -1$
we get points on the graph arbitrarily far away, while also arbitrarily close to, three of the four semi-axes. This cannot be the case for any reflection of a graph of a continuous bijective function on the reals, so there cannot be points off the axes, which correspond to the graphs of $f(x) = \phi x$ and $f(x) = -\phi^{-1}x$.