Show that $f(x)=cx$, $\forall x\in\mathbb{R}$.

For $t>0$ we have $tf(x)=\int_0^t(f(x+y)-f(y))dy.$ Differentiation with respect to $t$ gives:

$$(*) \quad f(x)=f(x+t)-f(t)$$

for all $x \in \mathbb R$ and all $t>0.$

With the continuity of $f$ we get, with $t \to 0+$: $f(x)=f(x)-f(0)$, hence $f(0)=0.$

The equation $f(x)=\frac 1 t \int_x^{x+t} f(y) dy-\frac 1 t \int_0^{t} f(y) dy$ shows that $f$ is differentiable.

From $(*)$ we see that for $t>0$ we have

$$ \frac{f(x+t)-f(x)}{t}=\frac{f(t)}{t}= \frac{f(t)-f(0)}{t}.$$

With $t \to 0+$ we derive $f'(x)=f'(0)$. Hence $f'$ is constant and the assertion follows.