calculusderivativesimplicit-differentiation
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
Since you know that the $x$ coordinate giving the right slope is $1$the point on the graph will be $(1,f(1))=(1,9)$.
The problem has solutions. Actually, all solutions satisfy the condition $y=x$. In fact if, in the equation $x^2y^2+xy=2$, you replace $y$ with $x$, you get $x^4+x^2=2$, which has two (and only two) real solutions: $\pm1$. So $\pm(1,1)$ are solutions of your problem. See the image below.
Best Answer
We have that
$$x^3-xy+y^3=0\implies 3x^2dx-dx-xdy+3y^2dy=0 \implies \frac{dy}{dx}=\frac{3x^2-y}{x-3y^2}=0$$
that is
$$3x^2=y \implies (x,y)=(t,3t^2)$$
$$x^3-xy+y^3=0 \iff t^3-3t^3+27t^6=0\iff t^3(27t^3-2)=0$$
that is
$$(x,y)=\left(\frac{\sqrt[3]2}{3},\sqrt[3]4\right)$$