a question on Tangent Lines and Implicit Functions

Question

Let $F(x,y)=x\sqrt{x^2+y^3}-2y+1$, a tangent to $F(x,y)=0$ curve at the point $(1,2)$ is needed.

The standard method requires $y'(1)$, which equals $-\frac{\frac{\partial F}{\partial x}(1,2)}{\frac{\partial F}{\partial y}(1,2)}$, but the denominator is $0$. What happens in this case?

Answer 1

$$\left(x\sqrt{x^2+y^3}-2y+1\right)'=0$$ or $$\sqrt{x^2+y^3}+\frac{x(2x+3y^2y')}{2\sqrt{x^2+y^3}}-2y'=0$$.

Let $m$ is a slope of our tangent.

Thus, $$\sqrt{1^2+2^3}+\frac{1\cdot(2\cdot1+3\cdot2^2m)}{2\sqrt{1^2+2^3}}-2m=0$$ or $$10=0,$$ which says that a slope does not exist, which gives the answer: $x=1$.