Unique solution of the differential equation with the initial value exercise:

Let $g$ be a continuous function on the domain $\mathbb{R}$ and suppose that for each $c$ the differential equation $y^{\prime}=g(y)$ has a unique solution, that satisfies the initial condition $y(0)=c$, and is defined on the domain $\mathbb{R}$. We can define the function $\phi(x, c)$ of the two variables $x$ and $c$, so that, as a function of $x$, it is the solution that satisfies the condition $y(0)=c$.
(a) Show that the translate of a solution of $y^{\prime}=g(y)$ is again a solution; that is, if $y=f(x)$ is a solution and $\alpha$ a number then $f(x-\alpha)$ is also a solution.
(b) Show that there is a unique solution that satisfies $y^{\prime}(\alpha)=c$, and that it is $\phi(x-\alpha, c)$
My draft work:
a) Let $y=f(x)$ it means $f'(x)=g(f(x))\Rightarrow f'(x-a)=g(f(x-a))\Rightarrow h'(s)=g(h(s))$ and $h(0)=c$
b)  for simplicity denote $h(s)=t$. $\,$ $h'(s)=g(h(s))\Rightarrow [h'(s)]'=g'(t)h'(s)$ from here if we get the derivative of $q(s)=\dfrac{h'(s)}{e^{g(h(s))}}\Rightarrow q'(s)=\dfrac{[h'(s)]'e^{g(h(s))}-[e^{g(h(s))}]'h'(s)}{(e^{g(h(s))})^2}=0$ we conclude that $h'(s)=le^{g(h(s))}$ for some number l.
However, I couldn't to show the uniqueness. Can anybody check my work if I am on the right track?? Please.
Thanks to @Martin R comments I am very suspicious about this may be typo. It should be $y(\alpha)=c$ not $y'(\alpha)=c$
 A: The proof of a) is correct, you should have written that $h(s) := f(x - \alpha)$ but I think it's okay to assume that this is what you meant by $h(s)$.
The proof of b) is not correct, first of all because
$$q'(s) = \frac{[h'(s)]' e^{g(h(s))} - [ e^{g(h(s))}]' h'(s) }{(e^{g(h(s))} )^2 } = \frac{e^{g(h(s))}[g'(t)h'(s) - g'(t)(h'(s))^2 ]}{(e^{g(h(s))})^2} \neq 0$$
In fact there are no reasons why $q'(s)$ should be $= 0$.
Also the proof is wrong because $g$ is supposed to be only continuos so you can't take the derivative of $g$ because $g$ could be non differentiable!!
This is the correct proof of b)
First of all there is a typo, it should be $y(\alpha) = c$ and not $y'(\alpha) = c$.
Let $h(s)$ be a solution of the problem
$$\begin{cases} h'(s) &= g(h(s)) \\ h(\alpha) &= c \end{cases} $$
Then, repeating your proof of a), one can show that $h(x+\alpha)$ is a proof of the following problem
$$\begin{cases} f'(x) &= g(f(x)) \\
f(0) &= c \end{cases} $$
So because this problem has, by hypothesis, a unique solution $f(x)$,one necessarily has that $h(x + \alpha) = f(x)$ therefore $h(x) = f(x - \alpha)$ and because by the definition $f(x) = \phi(x,c)$ you also have $h(x) = \phi(x - \alpha , c)$
Now about c) let $f(x) := \phi(x,c)$, so $f$ is the unique solution of the problem
$$\begin{cases} f'(x) &= g(f(x)) \\
f(0) &= c \end{cases}$$
Let $h(s) := f(s + t)$ then one clearly has $h(s) = \phi(s+t,c)$ so moreover $h(0) = \phi(t,c)$ and also differentiating respect to $s$ $h'(s) = f'(s+t) = g(f(s+t)) = g(h(s))$ therefore $h$ satisfies the problem
$\begin{cases} h'(s) &= g(h(s)) \\
h(0) &= \phi(t,c) \end{cases} $
So by definition $h(s) = \phi(s,\phi(t,c)$, therefore
$$\phi(s+t,c) = h(s) = \phi(s,\phi(t,c))$$
