Taking logarithms in (b), we obtain the identity
$$\log \phi\biggl(\frac{x}{2}\biggr) + \log \phi\biggl(\frac{x+1}{2}\biggr) = \log \phi(x) + \log d.\tag{1}$$
Differentiating $(1)$ twice, that becomes
$$\frac{1}{4}\Biggl(g\biggl(\frac{x}{2}\biggr) + g\biggl(\frac{x+1}{2}\biggr)\Biggr) = g(x).\tag{2}$$
So $g$ is a continuous function with period $1$ that satisfies the relation $(2)$. By periodicity, we can assume $g$ is defined on all of $\mathbb{R}.$
We want to show that $\phi$ is constant, so in particular that $g\equiv 0$. Choose $x_1,x_2 \in [0,1]$ so that
$$g(x_1) = \min \{ g(x) : x\in [0,1]\};\qquad g(x_2) = \max \{ g(x) : x\in [0,1]\}.$$
Since $g$ is continuous and $[0,1]$ is compact, such points exist. Since $g$ has period $1$, $g(x_1)$ is also the global minimum that $g$ attains on $\mathbb{R}$, and $g(x_2)$ the global maximum.
By $(2)$, we have
$$g(x_1) = \frac{1}{4}\Biggl(g\biggl(\frac{x_1}{2}\biggr) + g\biggl(\frac{x_1+1}{2}\biggr)\Biggr) \geqslant \frac{1}{4}\bigl(g(x_1) + g(x_1)\bigr) = \frac{1}{2}g(x_1),$$
so $g(x_1) \geqslant 0$. The same argument shows $g(x_2) \leqslant 0$, hence $g\equiv 0$, as desired.
Therefore, it follows that
$$\biggl(\frac{d}{dx}\log \phi\biggr)(x) \equiv a = \text{const},$$
and hence
$$\log \phi(x) = ax+b$$
and $\phi(x) = e^{ax+b}$ for some constants $a,b\in\mathbb{R}$. It remains to show that $a = 0$.
