How to find point in Descartes Folium with slope of -1/3? I stumbled upon a problem in my calculus book that asked to find the point in $x^3 + y^3 = 3xy$ that had a slope perpendicualr to $y = 3x + 1$ and also was in the first quadrant.
I began by getting the derivative of the equation and got $\dfrac{x^2-y}{x-y^2}$. 
Given that the slope is $-\frac13$, I wrote $\dfrac{x^2-y}{x-y^2}=-\frac13
$, which got me to $3x^2+x=y^2+3y$, and that is where I got stuck, because I cannot solve for $x$ or $y$, to substitute in the original equation.
I looked for similar examples on the internet, and only found some where the slope was $0$, which can be solved because $y = x^2$, which can be substituted back.
Hope this is redacted well enough, any help is really appreciated. 

Let's find a parametric equation of the curve defined by $x^3+y^3=3xy$.
Let $y=xt$, then
$$x^3(1+t^3)=3x^2t$$
Hence
$$x=\frac{3t}{1+t^3}$$
$$y=\frac{3t^2}{1+t^3}$$
Now, the tangent vector is given by
$$x'(t)=\frac{3(1+t^3)-3t(3t^2)}{(1+t^3)^2}=\frac{3-6t^3}{(1+t^3)^2}$$
$$y'(t)=\frac{6t(1+t^3)-3t^2(3t^2)}{(1+t^3)^2}=\frac{6t-3t^4}{(1+t^3)^2}$$
You want $t$ such that $y'(t)=-\frac13x'(t)$, that is
$$6t-3t^4=2t^3-1$$
You have thus a quartic equation to solve: $3t^4+2t^3-6t-1=0$. There does not seem to be a trivial solution, so you'll have to go through Descartes' or Ferrari's method to solve this, or a numerical method.
There are two real roots (hence two points), with numerically
$$t_1=-0.1678460300438111, x_1=-0.5059304363344097, y_1=0.0849184152170638$$
$$t_2=1.1301467382482941, x_2=1.3875575405757694, y_2=1.5681436286135308$$

