Show that $y/x$ tends to a finite limit as $x \to + \infty$ and determine this limit. 
Let $y=f(x)$ be that solution of the differential equation $$y' = \frac{2y^2+x}{3y^2+5}$$ which satisfies the initial condition $f(0)=0$. (Do not attempt to solve this differential equation.)
  (a) The diffeerential 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 \geq0$ 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 $x \to + \infty$. Give full details of your reasoning.
  (d) Show that $y/x$ tends to finite limit as $x \to + \infty$ and determine this limit.

a) From the equation I have $$y''= \frac{(2yy'+1)(3y^2+5)-(2y^2+x)(3yy')}{(3y^2+5)^2}=\frac{3y^2+5-x3yy'}{(3y^2+5)^2}$$ so $y''>0$ when $x=0$ ,$f$ has a relative minimum at 0.  
b) Let $a= \frac{2}{3},b=\frac{20}{9}$,we know $f(x)>0=ax-b$ when $x=\frac{10}{3}$.
Let $g(x)=f(x)-ax+b$ then $g'(x)=f'(x)-a \geq 0$ for $x \geq \frac{10}{3}$ . so $g(x)\geq g(\frac{10}{3})>0$ for $x \geq \frac{10}{3}$ .  It means $f(x) > ax-b$ for each $x\geq \frac{10}{3}$.
c) For those $x\geq 10/3$ I have $0< \frac{x}{y^2} <\frac{x}{(\frac{2}{3}x-\frac{20}{9})^2}$ as $x \to + \infty$.By the sandwiches theorem $\frac{x}{y^2} \to 0$ as $x \to + \infty$.
d) Notice $y > ax -b$ ,so $y \to + \infty$ as $ x \to + \infty$ the limit $\frac{y}{x}$ has the form $\frac{+ \infty}{+ \infty}$.It seems like L'Hopital's rule stuff.But the book only gives application and proof of LHR in the form $\frac{0}{0}$.So I write $y/x=\frac{1/x}{1/y}$ in the form $0/0$, use L'Hopital's rule I get $$\frac{y^2}{x^2} \frac{2+\frac{x}{y^2}}{3+\frac{5}{y^2}}$$ So I'm stuck.
Any help is appreciated.

Even though the book gives L'Hospital rule only for the case when $f(x)$ and $g(x)$ tend to $0$, the rule holds also for $\frac{\infty}{\infty}$ case, too.
Since $\lim_{x \to +\infty} \frac{y'}{x'} = \lim_{x \to +\infty} \frac{2y^2}{3y^2 + 5} + \lim_{x \to +\infty} \frac{x}{3y^2 + 5} = \frac{2}{3} + 0$, the answer to the part d) is $\frac{2}{3}$. The last equality is coming from composition theorem $3.5$, and the part c).
