You do not say where $f$ is defined, so let us assume it is defined on an open (bounded or unbounded) interval $D = (a,b)$. You can also take a half-open or a closed interval if you are willing to work with left-hand / right-hand derivatives in the boundary points $a$ or $b$.
Let us introduce the continuos function
$$g = \frac{1}{2} +\frac{1}{f^3} .$$
Note that $g > \frac{1}{2}$.
You claimed that $f'' = g$, but the proof breaks down in the points $\xi \in D$ where $f'(\xi) = 0$. However we shall prove the following:
Either $f = 1$ (in which case $f''(\xi) = 0 \ne g(\xi)$ for all $\xi$) or $f \ne 1$ and $f'' = g$.
We have $f'(x)^2 \ge 0$, hence $f(x) - \frac{1}{f(x)^2} \ge 0$ which means $f(x)^3 \ge 1$, that is, $f \ge 1$. Let $A = f^{-1}(1)$. This is a closed subset of $D$. We have $\xi \in A$ iff $f'(\xi) = 0$. Thus for all $\xi \in D \setminus A$ we correctly get
$$(*) \quad f''(\xi) = \frac{1}{2} +\frac{1}{f(\xi)^3} = g(\xi) .$$
What happens if $\xi \in A$?
Case 1. $\xi$ is not a limit point of $A$, that is, $\xi$ is an isolated point of $A$.
Then a pointed open neigborhood $U = (\xi - \varepsilon, \xi + \varepsilon) \setminus \{ \xi \}$ of $\xi$ is contained in $D \setminus A$. Now we apply Rolle's theorem. For $x \in U$ we have
$$ \frac{f'(x) - f'(\xi)}{x - \xi} = f''(\zeta(x)) = g(\zeta(x))$$
for some $\zeta(x)$ between $x$ and $\xi$. If $x \to \xi$, then clearly $\zeta(x) \to \xi$. Hence
$$f''(\xi) = \lim_{x \to \xi} \frac{f'(x) - f'(\xi)}{x - \xi} = \lim_{\zeta(x) \to \xi} g(\zeta(x)) = g(\xi) .$$
Thus $(*)$ is also correct for $\xi \in A$ being an isolated point of $A$.
Case 2. $\xi$ is a limit point of $A$, that is, $\xi \in A' =$ set of limit points of $A$.
Then we find a sequence $(x_n)$ in $A \setminus \{ \xi \}$ such that $x_n \to \xi$. Therefore
$$f''(\xi) = \lim_{n\to \infty}\frac{f'(x_n) - f'(\xi)}{x_n - \xi} = 0$$
because $f'(x_n) = f'(\xi) = 0$.
Summarizing we have $f''(\xi) = 0$ iff $\xi \in A'$ and $f''(\xi) = g(\xi) > \frac{1}{2}$ iff $\xi \notin A'$.
We claim that if $A' \ne \emptyset$, then $A = D$ which means $f = 1$. This is the alternative solution of Mohammad Riazi-Kermani's answer.
So pick any $\xi \in A'$ and let $C$ be the connected component of $\xi$ in $A$. Since $A$ is closed in $D$, also $C$ is closed in $D = (a,b)$. As a connected subset of the real line $C$ must be an interval with boundary points $c \le d$.
Assume $a < c$. Then $c \in C$ because $C$ is closed in $D$. Moreover $c \in A'$. This is clear for $c < d$ and also for $c = d$ (in this case we have $\xi \in [c,c]$, thus $c = \xi \in A'$). For each $0 < \varepsilon < c-a$ the interval $(c-\varepsilon,c)$ must contain a point of $D \setminus A$ (otherwise $C' = (c-\varepsilon,c) \cup C$ would be a connected subset of $A$, hence $C' \subset C$ which is a contradiction). Thus we can find a sequence $(x_n)$ in $D \setminus A$ such that $x_n < c$ and $x_n \to c$. Then again by Rolle's theorem
$$\frac{f'(x_n) - f'(c)}{x - c} = f''(\zeta(x_n))$$
for some $\zeta(x_n)$ between $x_n$ and $c$. Since $f'(x_n) - f'(c) = f'(x_n) \ne 0$, we see that $f''(\zeta(x_n)) \ne 0$ and conclude $\zeta(x_n) \notin A'$. Hence
$$\frac{f'(x_n) - f'(c)}{x - c} = g(\zeta(x_n) .$$
We have $\zeta(x_n) \to c$ as $n \to \infty$ so that
$$f''(c) = \lim_{n \to \infty} \frac{f'(x_n) - f'(c)}{x_n - c} = \lim_{n \to \infty} g(\zeta(x_n)) = g(c) \ne 0 .$$
But on the other hand $f''(c) = 0$ because $c \in A'$. This is a contradiction.
Therefore $a = c$.
Similarly we show that $b = d$. Hence $C = D$ which shows $A = D$ and $f = 1$.
If $f \ne 1$, then $A \ne D$. We have shown that if $A' \ne \emptyset$, then $A = D$, i.e. $f = 1$, which is impossible. Hence $A' = \emptyset$ and $f'' = g$.
Your solution may be simplified by noting that the system $$f(x) + g(1) = x \\ f(x) + g(0) = 0$$ implies that the difference $$f(x) + g(1) - f(x) - g(0) = g(1) - g(0) = x$$ for all $x$; therefore, $g$ is not well-defined. At this point, I would consider the reasoning valid.
As for Spivak's solution, he first establishes that $f(x) + g(0) = 0$ for all $x$ by choosing $y = 0$. That means that for any choice of $x$, we must have $f(x) = -g(0)$. this implies $f$ is a constant function. Upon substituting back into the original functional equation, $xy = f(x) + g(y) = -g(0) + g(y)$ for all $y$. This is allowed because $f(x)$ is constant for any $x$. At this point, we have already run into a problem because $g(0)$ is a constant, $g(y)$ doesn't depend on $x$, yet their difference $g(y) - g(0) = xy$. How can the LHS not depend on $x$, yet the RHS does? To formalize this in terms of a logical contradiction, Spivak takes the argument to completion: he then chooses $x = 0$, which then implies $$g(y) = g(0)$$ for all $y$. Then he finally returns to the functional equation $$f(x) + g(y) = xy,$$ armed with both $f(x) = -g(0)$ and $g(y) = g(0)$, and concludes $$xy = f(x) + g(y) = -g(0) + g(0) = 0$$ for all $x$ and $y$, but this is absurd since selecting $x = y = 1$ implies $1 = 0$.
The entire proof relies on a strategy that you yourself employed: to take the functional equation, set one variable to a particular value, and show that the resulting equation for the other variable, which is still valid for all values of that variable, results in a contradiction.
Best Answer
The given solution is wrong and you are right. It should be $$x(t) = -\frac{b(t)}{a(t)}$$