Strange point on a curve

Question

The curve $$x^3+(y^2+1)\,x^2-5x+\frac{2}{xy}=0$$ has a solution at $(x,y)=(1,1)$. Implicitly differentiating wrt $x$ gives $$3x^2+2x^2y\frac{dy}{dx}+2(y^2+1)\,x-5-\frac{2}{x^2 y}-\frac{2}{xy^2}\frac{dy}{dx}=0.$$ Naively plugging in $(x,y)=(1,1)$ gives $0=0$, but solving for $\frac{dy}{dx}$ gives $$\frac{dy}{dx}=\frac{2\frac{y}{x}+5xy^2-2x^2y^2(y^+1)-3x^3y^2}{2x^3y^3-2}$$ which has no limit at $(x,y)\rightarrow(1,1)$, as can be checking by e.g. setting $y=1$ and $y=x$ then taking $x\rightarrow 1$.

What's going on with the curve around $(1,1)$? Is the point pathological in some sense? Is it just an isolated solution?

Answer 1

The graph provided by Geogebra shows a concave "nose" at $(1,1,0)$.

Answer 2

I would say

$F_{xx}(1,1)=14, \quad F_{yy}(1,1)=6, \quad F_{xy}(1,1)=F_{yx}(1,1)=6$

$\Rightarrow F_{xx}F_{yy}-F_{xy}^2=14\cdot 6-6^2=\color{red}{48 > 0}$

$\Rightarrow $ according to smart books the point (1,1) is an isolated point of the curve.