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. 

I cannot understand what you mean "had a slope perpendicular to $y=3x+1$ ". 
A slope is an number and $y=3x+1$ is the equation of a straight line. A number perpendicular to a line is non-sens.
I suppose that you mean "to find the point where the tangent to $x^3 + y^3 = 3xy$ is  perpendicular to $y = 3x + 1$ "
If so, the slope of the tangent must be $-\frac13$.
Differentiating $x^3 + y^3 = 3xy$ leads to :
$$3x^2dx+3y^2dy=3ydx+3xdy$$ 
$$\frac{dy}{dx}=\frac{y-x^2}{y^2-x}$$
All the points $(x,y)$ where the slope of the tangent is $-\frac13$ are given by the system of equations :
$$\begin{cases}
x^3 + y^3 = 3xy \\
\frac{y-x^2}{y^2-x}=-\frac13
\end{cases}$$
Solve the system of equations for the points $(x,y)$ and select the point in the first quadrant. (Numerical solving).
In addition. The analytical solving leads to the exact formulas :


