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Find coordinate in first quadrant which tangent line to $x^3-xy+y^3=0$ has slope 0

First, I do implicit differentiation:

$frac{3x^2-y}{x-3y^2}=y'$

so I look at the numerator and go hmmm if i put in (1,3) that makes the slope 0.

But then I graph it on a software and i get the following image-

Clearly, this is an incorrect point. I did double check that the eqn i typed in was correct and that i did the implicit differentiation right

We have that

$$x^3-xy+y^3=0implies 3x^2dx-dx-xdy+3y^2dy=0 implies frac{dy}{dx}=frac{3x^2-y}{x-3y^2}=0$$

that is

$$3x^2=y implies (x,y)=(t,3t^2)$$

$$x^3-xy+y^3=0 iff t^3-3t^3+27t^6=0iff t^3(27t^3-2)=0$$

that is

$$(x,y)=left(frac{sqrt[3]2}{3},sqrt[3]4right)$$

You solved only half of the problem. You have that the derivative is $0$, but you also need to use the fact that the point is on the graph of your line. You have two equations with two unknowns. Since they are not linear equations, you might have multiple solutions.

Just plug in $y=3x^2$ into your original equation, and solve for $x$