Given the following equation:
$$ x^3y -y^3x -6 = 0 $$
determine using implicit differentiation $y'(2)$. (exact wording) Where $y' = \frac{\partial y}{\partial x}$
I called
$$ f(x,y) = x^3y -y^3x -6 = 0 $$ Here is what I have done so far:
$$ \begin{align*} \frac{\partial}{\partial x}(f(x,y)) &= 3x^2y - y^3 \\ \frac{\partial}{\partial y}(f(x,y)) &= x^3 - 3y^2x \\ f'(x) &= -\frac{\frac{\partial}{\partial x} (f(x,y))}{\frac{\partial}{\partial y} (f(x,y))} = \frac{y^3- 3x^2y}{x^3 - 3y^2x} \end{align*} $$
Here I am stuck because I cannot express y as a function of x:
$$ x^3y -y^3x -6 = 0 \\ $$
or at least i do not see a way. Without expressing $y = g(x)$ I can not solve $f'(x)$. Am i missing something here?
For $x=2$ we obtain: $$8y-2y^3-6=0$$ or $$y^3-4y+3=0$$ or $$y^3-y^2+y^2-y-3y+3=0$$ 0r $$(y-1)(y^2+y-3)=0,$$ which gives three points.
Can you end it now?
Also, I got $$y'=\frac{y^3-3x^2y}{x^3-3xy^2}.$$