Differential equation $y' = (2y^2 + x)/(3y^2 + 5)$ (Apostol, section 8.28, ex. 30) Problem
This is from Apostol's Calculus book, section 8.28, exercise 30.

Let $y = f(x)$ be that solution of the differential equation
$$y' = \dfrac{2y^2 + x}{3y^2 + 5}$$
which satisfies the initial condition $f(0) = 0$.
(a) The differential equation shows that $f'(0) = 0$. Discuss whether $f$ has a relative maximum or minimum or neither at 0.
(b) Notice that $f'(x) \geq 0$ for each $x \geq 0$ and that $f'(x) \geq \frac{2}{3}$ for each $x \geq \frac{10}{3}$. Exhibit two positive numbers $a$ and $b$ such that $f(x) > ax - b$ for each $x \geq \frac{10}{3}$.
(c) Show that $x/y^2 \to 0$ as $x \to +\infty$.
(d) Show that $y/x$ tends to a finite limit as $x \to +\infty$ and determine this limit.

Solution attempt
Please, I would like to ask for verification of my attempt below.
(a) The second derivative is given by:
$$y'' = \dfrac{(3y^2 + 5)(4yy' + 1) - 6yy'(2y^2 + x)}{(3y^2 + 5)^2}$$
At zero, the value is:
$$f''(0) = \dfrac{5}{(5)^2} = \dfrac{1}{5}$$
Since this value is positive, we can conclude that the concavity of $f$ at 0 is up, so $f$ has a relative minimum at 0.
(b) Since $f'(x) \geq \frac{2}{3}$ for each $x \geq \frac{10}{3}$, in order to have the line $ax - b$ below $f(x)$, we can choose $a = \frac{2}{3}$. Also, we can choose $b$ such that $ax - b = 0$ at $x = \frac{10}{3}$ (since $f(x) \geq 0$ there). This gives:
$$\frac{2}{3} \cdot \frac{10}{3} - b = 0 \implies b = \frac{20}{9}$$
So, the numbers are $a = \frac{2}{3}$ and $b = \frac{20}{9}$.
(c) In item (b), we found that:
$$y > \frac{2}{3}x - \frac{20}{9}$$
for $x \geq \frac{10}{3}$. This implies:
$$\frac{3}{2} y + \frac{10}{3} > x$$
for $x \geq \frac{10}{3}$. Dividing both sides by $y^2$, this becomes:
$$\dfrac{3}{2y} + \dfrac{10}{3y^2} > \dfrac{x}{y^2}$$
Also, we know that $x/y^2 \geq 0$, so we have:
$$\dfrac{3}{2y} + \dfrac{10}{3y^2} > x/y^2 \geq 0$$
Since we know that $y > \frac{2}{3}x - \frac{20}{9}$, we have that $y \to +\infty$ as $x \to +\infty$, so the left-hand side of the left-hand inequality above tends to zero as $x \to \infty$:
$$\dfrac{3}{2y} + \dfrac{10}{3y^2} \to 0\text{ as }x \to \infty$$
Therefore, by the squeeze theorem, it follows that $x/y^2 \to 0$ as $x \to +\infty$.
(d) From item (c), we know that $x/y^2 \to 0$ as $x \to +\infty$. So:
$$\begin{aligned}
    \lim_{x \to +\infty} y' &= \lim_{x \to +\infty} \dfrac{2y^2 + x}{3y^2 + 5} \\
    &= \lim_{x \to +\infty} \dfrac{2 + x/y^2}{3 + 5/y^2} \\
    &= \dfrac{2 + 0}{3 + 0} \\
    &= \dfrac{2}{3}
\end{aligned}$$
So, as $x \to +\infty$, the function $y$ approaches a line of slope $\frac{2}{3}$. That is, $y$ approaches a line of the form $y=\frac{2}{3}x + C$, so $y/x \to \frac{2}{3}$ as $x \to +\infty$.
The argument in part (d) seems a bit informal; I'm not sure how to make it more rigorous.

Based on Paul Sinclair's suggestion in the comments to the question, here is an attempt at a new argument for part (d).
We found that $\lim_{x \to +\infty} y' = 2/3$. This means that, given $\epsilon > 0$, there is a $N > 0$ such that, if $x > N$, then
$$\left|y' - \dfrac{2}{3}\right| < \epsilon$$
This implies
$$-\epsilon < y' - \dfrac{2}{3} < \epsilon \implies -\epsilon + \dfrac{2}{3} < y' < \epsilon + \dfrac{2}{3}$$
Integrating, we get:
$$\begin{aligned}
    &\int_0^x \left(-\epsilon + \dfrac{2}{3}\right) dz < \int_0^x y' dz < \int_0^x \left(\epsilon + \dfrac{2}{3}\right) dz \\
    \implies & \left(-\epsilon + \dfrac{2}{3}\right)x < y - y(0) < \left(\epsilon + \dfrac{2}{3}\right)x \\
    \implies & \left(-\epsilon + \dfrac{2}{3}\right)x < y < \left(\epsilon + \dfrac{2}{3}\right)x
\end{aligned}$$
Dividing by $x$ (since $x \neq 0$), we get:
$$\begin{aligned}
    &-\epsilon + \dfrac{2}{3} < \dfrac{y}{x} < \epsilon + \dfrac{2}{3} \\
    \implies & -\epsilon < \dfrac{y}{x} - \dfrac{2}{3} < \epsilon \\
    \implies & \left|\dfrac{y}{x} - \dfrac{2}{3}\right| < \epsilon
\end{aligned}$$
That is, we have shown that, given $\epsilon > 0$, there is a $N > 0$ such that, if $x > N$, then $|y/x - 2/3| < \epsilon$. That is,
$$\lim_{x \to +\infty} \dfrac{y}{x} = \dfrac{2}{3}$$
as required.
