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Let's have an equation defining Folium of Descartes $$ x^3 + y^3 - 3axy = 0, \quad a>0. $$ If we will divide both sides of it by $x^3$ we will get $$ \left(\frac{y}{x}\right)^3 = 3a \frac{y}{x} \cdot \frac{1}{x} -1. $$ Then, my book tells me that from the above equation we can conclude that for $|x|>3a$ the ratio $|y/x|$ is bounded. It seems like that should be something obvious but I can't see it. Can anybody explain it to me?

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Here's with assuming $x>0.$ Then $k=3a/x<1$ given $|x|=x>3a,$ so if one puts $u=y/x$ one gets $u^3=uk-1,$ then $u^3

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    I'm sorry, why $u^3 < uk$? Shouldn't it be $u^3 < u-1$ if $k<1$? – 2017-02-15
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    If $u^3=(3a/x)(u)-1$ [because $u=y/x$] then $u^3<(3a/x)u$ [because $3a/x-1<3a/x$, then $u^3$k=3a/x.$] – 2017-02-15
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    That's right, but how can you divide by $|u|$ and write $\frac{u^3}{|u|} = |u|^2$? Don't you need any assumption about the sign of $u$? – 2017-02-16
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    The question is to bound $|y/x|.$ So just let $u=|y/x|$ so that $u>0.$ So in this redefinition of $u,$ I only use that $u^3/u=u^3/|u|=|u|^3/|u|=|u|^2.$ I should have defined $u$ this way in the first place. – 2017-02-17